11-6 Secants, Tangents, and Angle Measures

Home , Tangent

Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

11-6 Secants, Tangents, and Angle Measures

Find each measure. Assume that segments that 4. appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc QTS is 248. So, the measure of arc TS is 144.

3. 6.

SOLUTION:

So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of 4. the angle the ramp makes with the ground? eSolutions Manual - Powered by Cognero Page 1

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc 5. has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

Therefore, the measure of the angle the ramp makes SOLUTION: with the ground is 15. Find each measure. Assume that segments that appear to be tangent are tangent. 8.

So, the measure of arc QTS is 248.

SOLUTION:

9. So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground? SOLUTION:

10. SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc

with the same endpoints, so its measure is 360 – 165 or 195. SOLUTION:

and ∠JMK ∠HMJ form a linear pair. Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 11. 8.

SOLUTION: SOLUTION:

Therefore, the measure of arc RQ is 28.

12. 9.

SOLUTION: SOLUTION:

13. 10.

SOLUTION:

So, the measure of arc PM is 144.

∠JMK and ∠HMJ form a linear pair. 14.

11.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

Therefore, the measure of arc RQ is 28. 15.

12.

SOLUTION:

SOLUTION: By Theorem 10.13, .

13. 16.

SOLUTION:

So, the measure of arc PM is 144. By Theorem 11.13, .

14. Solve for .

We know that . Substitute.

SOLUTION: 17. SPORTS The multi-sport field shown includes a Arc BD and arc BCD are a minor and major arc that softball field and a soccer field. If , share the same endpoints. find each measure.

15.

a. b. SOLUTION:

a. By Theorem 11.13, . SOLUTION: Substitute.

By Theorem 10.13, . Simplify.

b. By Theorem 11.14,

. Substitute. 16.

Simplify.

CCSS STRUCTURE Find each measure.

18. SOLUTION:

By Theorem 11.13, .

Solve for .

We know that . Substitute. SOLUTION:

By Theorem 11.14, . Substitute. 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , Simplify. find each measure.

19.

a. b. SOLUTION:

a. By Theorem 11.13, .

Substitute. SOLUTION:

By Theorem 11.14, . Simplify. Substitute.

b. By Theorem 11.14,

. Simplify.

Substitute.

Simplify.

20. m(arc JNM)

CCSS STRUCTURE Find each measure. 18.

SOLUTION: SOLUTION: By Theorem 11.14, . Substitute.

So, the measure of arc JNM is 205. Simplify.

21.

19.

SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify. By Theorem 11.14, . Substitute.

22. Simplify.

SOLUTION:

20. m(arc JNM) By Theorem 11.14, . Substitute.

Simplify.

23.

SOLUTION:

SOLUTION: So, the measure of arc JNM is 205. By Theorem 11.14, .

Substitute. 21.

Solve for .

24. JEWELRY In the circular necklace shown, A and B SOLUTION: are tangent points. If x = 260, what is y?

By Theorem 11.14, . Substitute.

Simplify.

22. SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify. By Theorem 11.14, . Substitute.

25. SPACE A satellite orbits above Earth’s equator. Simplify. Find x, the measure of the planet’s arc, that is visible to the satellite.

23.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x. SOLUTION:

By Theorem 11.14, . Substitute. Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

Solve for . ALGEBRA Find the value of x.

24. JEWELRY In the circular necklace shown, A and B 26. are tangent points. If x = 260, what is y? SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

27. SOLUTION:

25. SPACE A satellite orbits above Earth’s equator. By Theorem 11.14, . Find x, the measure of the planet’s arc, that is visible to the satellite. Solve for .

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

28. SOLUTION:

Therefore, the measure of the planet’s arc that is By Theorem 11.14, . visible to the satellite is 168. Solve for . ALGEBRA Find the value of x.

26. SOLUTION:

By Theorem 11.14, . Solve for .

29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

27. SOLUTION:

By Theorem 11.14, . SOLUTION: Solve for . a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

b. Let x be the measure of the camera’s viewing 28. angle.

SOLUTION: By Theorem 11.14, .

By Theorem 11.14, . Solve for .

Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove:

29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the

arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in SOLUTION: the photograph, what viewing angle should be used? Statements (Reasons) 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

SOLUTION: 5. (Subtraction Prop.) a. Let x be the measure of the carousel that appears

in the shot. 6. (Distributive Prop.)

By Theorem 11.14, . Solve for . 31. Case 2 Given: tangent and secant

Prove:

b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . SOLUTION: Solve for . Statements (Reasons)

1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The

meas. of an inscribed the measure of its CCSS ARGUMENTS For each case of

Theorem 11.14, write a two-column proof. intercepted arc.) 30. Case 1 3. (Exterior

Given: secants and Theorem)

Prove: 4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

32. Case 3 SOLUTION: Given: tangent and Statements (Reasons)

1. and are secants to the circle. (Given) Prove: 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior

Theorem) SOLUTION: 4. (Substitution) Statements (Reasons) 1. and are tangents to the circle. (Given) 5. (Subtraction Prop.) 2. (The meas.

6. (Distributive Prop.) of an secant-tangent the measure of its

31. Case 2 intercepted arc.) 3. (Exterior Given: tangent and secant Theorem) Prove: 4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

SOLUTION: 33. PROOF Write a paragraph proof of Theorem 11.13. Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given)

2. , ( The

meas. of an inscribed the measure of its

intercepted arc.) 3. (Exterior Theorem)

4. (Substitution) a. Given: is a tangent of . is a secant of .

5. (Subtraction Prop.) ∠CAE is acute.

Prove: m∠CAE = m(arc CA) 6. (Distributive Prop.) b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA) 32. Case 3

Given: tangent and SOLUTION: Prove: a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = SOLUTION: m(arc FC) + m(arc CA). So, 90 = m(arc FC) + Statements (Reasons) (arc CA) by Division Prop., and m∠FAC + m∠CAE 1. and are tangents to the circle. (Given) = m(arc FC) + m(arc CA) by substitution. 2. (The meas. m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m of an secant-tangent the measure of its (arc FC) + m(arc CA). By Subtraction Prop., intercepted arc.)

3. (Exterior m∠CAE = m(arc CA). Theorem) b. Proof: Using the Angle and Arc Addition 4. (Substitution) Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since 5. (Subtraction Prop.) and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a 6. (Distributive Prop.) measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since 33. PROOF Write a paragraph proof of Theorem 11.13. ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m∠CAB = m(arc

CF) + m(arc CDA) – m(arc CF). Then by subtraction, m∠CAB = m(arc CDA).

a. Given: is a tangent of . is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA)

b. Prove that if ∠CAB is obtuse, m∠CAB = m 34. WALLPAPER The design shown is an example of (arc CDA) optical wallpaper. is a diameter of If m∠A

= 26 and = 67, what is SOLUTION: Refer to the image on page 766. a. Proof: By Theorem 11.10, . So, ∠FAE SOLUTION: is a right ∠ with measure of 90 and arc FCA is a First, find the measure of arc BD. semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). Since is a diameter, arc BDE is a semicircle and By substitution, 90 = m∠FAC + m∠CAE and 180 = has a measure of 180. m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution. 35. MULTIPLE REPRESENTATIONS In this m∠FAC = m(arc FC) since ∠FAC is inscribed, so problem, you will explore the relationship between Theorems 11.12 and 11.6. substitution yields m(arc FC) + m∠CAE = m a. GEOMETRIC Copy the figure shown. Then (arc FC) + m(arc CA). By Subtraction Prop., draw three successive figures in which the position of point D moves closer to point C, but points A, B, m∠CAE = m(arc CA). and C remain fixed.

b. Proof: Using the Angle and Arc Addition b. TABULAR Estimate the measure of for Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar each successive circle, recording the measures of CDA) = m(arc CF) + m(arc FDA). Since and in a table. Then calculate and record and is a diameter, ∠FAB is a right angle with a the value of x for each circle. measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF c. VERBAL Describe the relationship between + 90 and m(arc CDA) = m(arc CF) + 180. Since and the value of x as approaches zero. ∠CAF is inscribed, m∠CAF = m(arc CF) and by What type of angle does ∠AEB become when substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and CF) = 90. By substituting for 90, m∠CAB = m(arc 11.6 described in part c. CF) + m(arc CDA) – m(arc CF). Then by subtraction, m∠CAB = m(arc CDA).

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. 34. WALLPAPER The design shown is an example of Then draw chords and . optical wallpaper. is a diameter of If m∠A = 26 and = 67, what is Refer to the image on page 766. b. SOLUTION: First, find the measure of arc BD.

SOLUTION: b. TABULAR Estimate the measure of for Find the difference of the two intercepted arcs and each successive circle, recording the measures of divide by 2. and in a table. Then calculate and record the value of x for each circle. 37. CHALLENGE The circles below are concentric. What is x? c. VERBAL Describe the relationship between and the value of x as approaches zero. What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and SOLUTION: 11.6 described in part c.

Use the arcs intercepted on the smaller circle to find SOLUTION: m∠A. a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and .

Use the arcs intercepted on the larger circle and m∠A to find the value of x. b.

38. REASONS Isosceles is inscribed in What can you conclude about and

Explain. c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle.

SOLUTION: 36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a Sample answer: m∠BAC = m∠BCA tangent that intersect outside a circle. because the triangle is isosceles. Since ∠BAC and SOLUTION: ∠BCA are inscribed angles, by Theorem 11.6, m Find the difference of the two intercepted arcs and (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. divide by 2. So, m(arc AB) = m(arc BC).

37. CHALLENGE The circles below are concentric. What is x? 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

SOLUTION:

SOLUTION: a. ; for all values except when at G, then .

Use the arcs intercepted on the smaller circle to find b. Because a diameter is m∠A. involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer.

40. OPEN ENDED Draw a circle and two tangents Use the arcs intercepted on the larger circle and that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures m∠A to find the value of x. of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer: 38. REASONS Isosceles is inscribed in What can you conclude about and Explain.

By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130,

and y (major arc) = 360° – 130° or 230°. SOLUTION: Sample answer: m∠BAC = m∠BCA 41. WRITING IN MATH A circle is inscribed within

because the triangle is isosceles. Since ∠BAC and If m∠P = 50 and m∠Q = 60, describe how ∠BCA are inscribed angles, by Theorem 11.6, m to find the measures of the three minor arcs formed by the points of tangency. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC). SOLUTION:

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ Sample answer: Use Theorem 11.14 to find each and KH. Explain. minor arc.

SOLUTION: a. ; for all values except when at G, then . The sum of the angles of a triangle is 180, so m∠R = b. Because a diameter is 180 – (50 + 60) or 70. involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to

the answer. Therefore, the measures of the three minor arcs are 130, 120, and 110. 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to 42. What is the value of x if and measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer:

A 23° B 31° C 64° D 128° By Theorem 11.13, So, 50° = SOLUTION:

Therefore, x (minor arc) = 130, By Theorem 11.14, . and y (major arc) = 360° – 130° or 230°. Substitute.

41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how Simplify. to find the measures of the three minor arcs formed by the points of tangency. SOLUTION:

So, the correct choice is C. Sample answer: Use Theorem 11.14 to find each minor arc. 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) J (1, 5)

SOLUTION: The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70. Here, C is the midpoint of . Use the midpoint formula, , to find the coordinates of C. Therefore, the measures of the three minor arcs are 130, 120, and 110.

42. What is the value of x if and

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

A 23° B 31° C 64° D 128° SOLUTION: SOLUTION: Since . By Theorem 11.14, . We know that vertical angles are congruent. Substitute. So, . Since , . Simplify. We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions? So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? 2 F (2, 10) A 64π units G (10, –6) B 32π units2 2 H (5, –3) C 12π units J (1, 5) D 8π units2 2 SOLUTION: D 2π units

Here, C is the midpoint of . SOLUTION: Use the midpoint formula, First, use the circumference to find the radius of the , to find the coordinates of circle. C.

So, the correct choice is J. The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. 44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

The area of the shaded regions is 32π. SOLUTION: So, the correct choice is B.

Since . Find x. Assume that segments that appear to be tangent are tangent. We know that vertical angles are congruent. So, . Since , . We know that the sum of the measures of all interior 46. angles of a triangle is 180. SOLUTION:

By Theorem 11.10, . So, is a right

Substitute. triangle. Use the Pythagorean Theorem.

Since , . Substitute.

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

2 A 64π units B 32π units2 2 C 12π units 47. 2 D 8π units SOLUTION: 2 D 2π units If two segments from the same exterior point are tangent to a circle, then they are congruent. SOLUTION: First, use the circumference to find the radius of the circle.

48. SOLUTION: The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x.

The area of the shaded regions is 32π. So, the correct choice is B. 49. PROOF Write a two-column proof. Given: is a semicircle; Find x. Assume that segments that appear to be tangent are tangent. Prove:

46. SOLUTION: By Theorem 11.10, . So, is a right SOLUTION: triangle. Given: is a semicircle. Use the Pythagorean Theorem.

Substitute. Prove:

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 47. 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) SOLUTION: 3. is a right angle (Def. of ⊥ lines) If two segments from the same exterior point are 4. (All rt. angles are .) tangent to a circle, then they are congruent.

5. (Reflexive Prop.) 6. (AA Sim.)

7. (Def. of

COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a 48. degree by using the Distance Formula and an SOLUTION: inverse trigonometric ratio. Use Theorem 11.10 and the Pythagorean Theorem. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) Solve for x. SOLUTION: Use the Distance Formula to find the length of each side.

49. PROOF Write a two-column proof. Given: is a semicircle;

Prove: Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

SOLUTION:

Given: is a semicircle. If

Use a calculator.

Prove:

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Use the Distance Formula to find the length of each side.

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) In right triangle XYZ, ZX is the length of the 5. (Reflexive Prop.) hypotenuse and YZ is the length of the leg opposite 6. (AA Sim.) Write an equation using the sine ratio. 7. (Def. of

COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a If degree by using the Distance Formula and an inverse trigonometric ratio. Use a calculator. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) Solve each equation. SOLUTION: 52. x2 + 13x = –36 Use the Distance Formula to find the length of each side. SOLUTION:

Therefore, the solution is –9, –4.

2 Since , triangle BCD is a right 53. x – 6x = –9 triangle. SOLUTION: So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

Therefore, the solution is 3. If 2 Use a calculator. 54. 3x + 15x = 0 SOLUTION:

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Therefore, the solution is –5, 0. Use the Distance Formula to find the length of each side. 2 55. 28 = x + 3x

SOLUTION:

In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite Therefore, the solution is –7, 4.

Write an equation using the sine ratio. 56. x2 + 12x + 36 = 0

SOLUTION:

Use a calculator. Therefore, the solution is –6.

Solve each equation. 52. x2 + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

Find each measure. Assume that segments that appear to be tangent are tangent.

1. 4.

SOLUTION: SOLUTION:

2. 5.

SOLUTION: SOLUTION:

So, the measure of arc TS is 144.

3. So, the measure of arc QTS is 248.

SOLUTION: SOLUTION:

So, the measure of arc LP is 150.

4. 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

5. Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that So, the measure of arc QTS is 248. appear to be tangent are tangent. 8. 6.

SOLUTION: SOLUTION:

11-6 Secants, Tangents, and Angle Measures So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several 9. barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes 10. with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15. ∠JMK and ∠HMJ form a linear pair. Find each measure. Assume that segments that appear to be tangent are tangent. 8.

11.

SOLUTION:

9. Therefore, the measure of arc RQ is 28. 12.

eSolutions Manual - Powered by Cognero Page 2

SOLUTION:

10.

13.

SOLUTION: SOLUTION:

∠JMK and ∠HMJ form a linear pair. So, the measure of arc PM is 144.

14. 11.

SOLUTION: Arc BD and arc BCD are a minor and major arc that SOLUTION: share the same endpoints.

Therefore, the measure of arc RQ is 28.

12. 15.

SOLUTION: SOLUTION:

By Theorem 10.13, .

13.

16.

SOLUTION:

So, the measure of arc PM is 144. SOLUTION: 14.

By Theorem 11.13, .

Solve for .

We know that . Substitute. SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints. SPORTS 17. The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

15.

a. b. SOLUTION: SOLUTION:

By Theorem 10.13, . a. By Theorem 11.13, . Substitute.

Simplify.

b. By Theorem 11.14,

16. . Substitute.

Simplify.

SOLUTION: CCSS STRUCTURE Find each measure. By Theorem 11.13, . 18. Solve for .

We know that . Substitute.

SOLUTION: 17. SPORTS The multi-sport field shown includes a By Theorem 11.14, . softball field and a soccer field. If , find each measure. Substitute.

Simplify.

a. 19. b. SOLUTION:

a. By Theorem 11.13, . Substitute.

Simplify. SOLUTION:

b. By Theorem 11.14, By Theorem 11.14, .

. Substitute.

Substitute.

Simplify.

CCSS STRUCTURE Find each measure. 18. 20. m(arc JNM)

SOLUTION:

By Theorem 11.14, . Substitute. SOLUTION:

Simplify.

So, the measure of arc JNM is 205.

21.

19.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute. By Theorem 11.14, . Substitute. Simplify.

Simplify.

22.

20. m(arc JNM) SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION: 23.

So, the measure of arc JNM is 205.

SOLUTION: 21. By Theorem 11.14, . Substitute.

Solve for .

SOLUTION:

By Theorem 11.14, . 24. JEWELRY In the circular necklace shown, A and B

Substitute. are tangent points. If x = 260, what is y?

Simplify.

22.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute. By Theorem 11.14, . Substitute. Simplify.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite. 23.

SOLUTION: SOLUTION: The measure of the visible arc is x and the measure By Theorem 11.14, . of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x. Substitute.

Solve for . Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x. 24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

26. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. 27. Find x, the measure of the planet’s arc, that is visible to the satellite. SOLUTION: By Theorem 11.14, . Solve for .

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is 28. visible to the satellite is 168. SOLUTION: ALGEBRA Find the value of x. By Theorem 11.14, . Solve for .

26. SOLUTION:

By Theorem 11.14, . Solve for .

27. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

28. SOLUTION: b. Let x be the measure of the camera’s viewing By Theorem 11.14, . angle.

Solve for . By Theorem 11.14, . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1

29. PHOTOGRAPHY A photographer frames a Given: secants and

carousel in his camera shot as shown so that the lines Prove: of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

SOLUTION: Statements (Reasons) 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior SOLUTION: Theorem)

a. Let x be the measure of the carousel that appears 4. (Substitution) in the shot.

By Theorem 11.14, . 5. (Subtraction Prop.) . Solve for 6. (Distributive Prop.)

31. Case 2 Given: tangent and secant

Prove: b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . Solve for .

SOLUTION: Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given) CCSS ARGUMENTS For each case of 2. , ( The Theorem 11.14, write a two-column proof. 30. Case 1 meas. of an inscribed the measure of its Given: secants and intercepted arc.) Prove: 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.)

SOLUTION: 6. (Distributive Prop.) Statements (Reasons)

1. and are secants to the circle. (Given) 32. Case 3 2. , (The Given: tangent and

Prove: measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) SOLUTION: Statements (Reasons) 6. (Distributive Prop.) 1. and are tangents to the circle. (Given)

2. (The meas. 31. Case 2

Given: tangent and secant of an secant-tangent the measure of its

Prove: intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) SOLUTION: Statements (Reasons) 6. (Distributive Prop.) 1. is a tangent to the circle and is a secant to the circle. (Given) 33. PROOF Write a paragraph proof of Theorem 11.13.

2. , ( The

meas. of an inscribed the measure of its

intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) a. Given: is a tangent of .

6. (Distributive Prop.) is a secant of . ∠CAE is acute. 32. Case 3 Prove: m∠CAE = m(arc CA)

Given: tangent and b. Prove that if ∠CAB is obtuse, m∠CAB = m

Prove: (arc CDA)

SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle SOLUTION: and Arc Addition Postulates, m∠FAE = m∠FAC + Statements (Reasons) m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m FAC + m CAE and 180 = 1. and are tangents to the circle. (Given) ∠ ∠ m(arc FC) + m(arc CA). So, 90 = m(arc FC) + 2. (The meas. (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution. of an secant-tangent the measure of its m∠FAC = m(arc FC) since ∠FAC is inscribed, so intercepted arc.) 3. (Exterior substitution yields m(arc FC) + m∠CAE = m

Theorem) (arc FC) + m(arc CA). By Subtraction Prop.,

4. (Substitution) m∠CAE = m(arc CA).

5. (Subtraction Prop.) b. Proof: Using the Angle and Arc Addition Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar 6. (Distributive Prop.) CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a 33. PROOF Write a paragraph proof of Theorem 11.13. measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m∠CAB = m(arc

CF) + m(arc CDA) – m(arc CF). Then by a. Given: is a tangent of . subtraction, m∠CAB = m(arc CDA). is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA)

b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA)

WALLPAPER SOLUTION: 34. The design shown is an example of optical wallpaper. is a diameter of If m∠A a. Proof: By Theorem 11.10, . So, ∠FAE

is a right ∠ with measure of 90 and arc FCA is a = 26 and = 67, what is semicircle with measure of 180. Since ∠CAE is Refer to the image on page 766. acute, C is in the interior of ∠FAE, so by the Angle SOLUTION: and Arc Addition Postulates, m FAE = m FAC + ∠ ∠ First, find the measure of arc BD. m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + Since is a diameter, arc BDE is a semicircle and (arc CA) by Division Prop., and m∠FAC + m∠CAE has a measure of 180. = m(arc FC) + m(arc CA) by substitution.

m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m CAE = m ∠ 35. MULTIPLE REPRESENTATIONS In this (arc FC) + m(arc CA). By Subtraction Prop., problem, you will explore the relationship between Theorems 11.12 and 11.6. m∠CAE = m(arc CA). a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position b. Proof: Using the Angle and Arc Addition of point D moves closer to point C, but points A, B, Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar and C remain fixed. CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a b. TABULAR Estimate the measure of for measure of 90 and arc FDA is a semicircle with a each successive circle, recording the measures of measure of 180. By substitution, m∠CAB = m∠CAF and in a table. Then calculate and record + 90 and m(arc CDA) = m(arc CF) + 180. Since the value of x for each circle. ∠CAF is inscribed, m∠CAF = m(arc CF) and by c. VERBAL Describe the relationship between substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc and the value of x as approaches zero. What type of angle does AEB become when Addition equation yields m(arc CDA) – m(arc ∠

CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and subtraction, m∠CAB = m(arc CDA). 11.6 described in part c.

34. WALLPAPER The design shown is an example of SOLUTION: a. Redraw the circle with points A, B, and C in the optical wallpaper. is a diameter of If m∠A same place but place point D closer to C each time.

= 26 and = 67, what is Then draw chords and . Refer to the image on page 766. SOLUTION: First, find the measure of arc BD.

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

35. MULTIPLE REPRESENTATIONS In this c. As the measure of gets closer to 0, the problem, you will explore the relationship between measure of x approaches half of Theorems 11.12 and 11.6.

a. GEOMETRIC Copy the figure shown. Then becomes an inscribed angle. draw three successive figures in which the position of point D moves closer to point C, but points A, B, d. and C remain fixed. 36. WRITING IN MATH Explain how to find the b. TABULAR Estimate the measure of for measure of an angle formed by a secant and a each successive circle, recording the measures of tangent that intersect outside a circle. and in a table. Then calculate and record SOLUTION: the value of x for each circle. Find the difference of the two intercepted arcs and divide by 2. c. VERBAL Describe the relationship between CHALLENGE and the value of x as approaches zero. 37. The circles below are concentric. What is x? What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c.

SOLUTION:

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Use the arcs intercepted on the smaller circle to find Then draw chords and . m∠A.

b. Use the arcs intercepted on the larger circle and m∠A to find the value of x.

c. As the measure of gets closer to 0, the 38. REASONS Isosceles is inscribed in measure of x approaches half of What can you conclude about and

becomes an inscribed angle. Explain.

36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a

tangent that intersect outside a circle. SOLUTION: SOLUTION: Find the difference of the two intercepted arcs and Sample answer: m∠BAC = m∠BCA divide by 2. because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m 37. CHALLENGE The circles below are concentric. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. What is x? So, m(arc AB) = m(arc BC).

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m G = 34, find the measures of minor arcs HJ SOLUTION: ∠ and KH. Explain.

SOLUTION: Use the arcs intercepted on the smaller circle to find a. ; for all values except when m∠A. at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to

Use the arcs intercepted on the larger circle and the answer. m∠A to find the value of x. 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. REASONS 38. Isosceles is inscribed in SOLUTION: What can you conclude about and Sample answer: Explain.

By Theorem 11.13, So, 50° = SOLUTION: Therefore, x (minor arc) = 130, Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and and y (major arc) = 360° – 130° or 230°. ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. 41. WRITING IN MATH A circle is inscribed within So, m(arc AB) = m(arc BC). If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. 39. CCSS ARGUMENTS In the figure, is a SOLUTION: diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

Sample answer: Use Theorem 11.14 to find each minor arc.

SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and The sum of the angles of a triangle is 180, so m∠R = x degrees. Hence solving leads to 180 – (50 + 60) or 70. the answer.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures Therefore, the measures of the three minor arcs are of the minor and major arcs formed. Explain your 130, 120, and 110. reasoning. 42. What is the value of x if and SOLUTION: Sample answer:

A 23° By Theorem 11.13, So, 50° = B 31° C 64° Therefore, x (minor arc) = 130, D 128° and y (major arc) = 360° – 130° or 230°. SOLUTION:

41. WRITING IN MATH A circle is inscribed within By Theorem 11.14, . If m∠P = 50 and m∠Q = 60, describe how Substitute. to find the measures of the three minor arcs formed by the points of tangency. SOLUTION: Simplify.

Sample answer: Use Theorem 11.14 to find each minor arc. So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) The sum of the angles of a triangle is 180, so m∠R = H (5, –3)

180 – (50 + 60) or 70. J (1, 5) SOLUTION: Here, C is the midpoint of .

Therefore, the measures of the three minor arcs are Use the midpoint formula, 130, 120, and 110. , to find the coordinates of C. 42. What is the value of x if and

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and A 23° what is m∠BAC? B 31° C 64° D 128° SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute. Since . We know that vertical angles are congruent. Simplify. So, . Since , . We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute.

Since , . So, the correct choice is C. 45. SAT/ACT If the circumference of the circle below 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on is 16π units, what is the total area of the shaded regions? circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) 2 J (1, 5) A 64π units 2 SOLUTION: B 32π units C 2 Here, C is the midpoint of . 12π units 2 Use the midpoint formula, D 8π units 2 , to find the coordinates of D 2π units C. SOLUTION: First, use the circumference to find the radius of the circle.

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC? The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

SOLUTION:

Since . The area of the shaded regions is 32π. We know that vertical angles are congruent. So, the correct choice is B. So, . Find x. Assume that segments that appear to be Since , tangent are tangent. . We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute. 46.

SOLUTION: By Theorem 11.10, . So, is a right , . Since triangle. Use the Pythagorean Theorem. 45. SAT/ACT If the circumference of the circle below

is 16π units, what is the total area of the shaded regions? Substitute.

2 A 64π units B 32π units2 2 C 12π units D 8π units2 2 D 2π units 47. SOLUTION: SOLUTION: First, use the circumference to find the radius of the If two segments from the same exterior point are circle. tangent to a circle, then they are congruent.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. 48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x.

The area of the shaded regions is 32π. So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent. 49. PROOF Write a two-column proof. Given: is a semicircle;

Prove: 46. SOLUTION: By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem. SOLUTION: Substitute. Given: is a semicircle.

Prove:

47. Proof: Statements (Reasons) SOLUTION: 1. MHT is a semicircle; (Given) If two segments from the same exterior point are 2. is a right angle. (If an inscribed angle tangent to a circle, then they are congruent. intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.)

7. (Def. of 48. SOLUTION: COORDINATE GEOMETRY Find the Use Theorem 11.10 and the Pythagorean Theorem. measure of each angle to the nearest tenth of a degree by using the Distance Formula and an Solve for x. inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each side.

49. PROOF Write a two-column proof.

Given: is a semicircle;

Prove:

Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to SOLUTION: Write an equation using the cosine ratio. Given: is a semicircle.

Prove: If

Use a calculator.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2)

Proof: SOLUTION: Statements (Reasons) Use the Distance Formula to find the length of each 1. MHT is a semicircle; (Given) side. 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.) In right triangle XYZ, ZX is the length of the 7. (Def. of hypotenuse and YZ is the length of the leg opposite

COORDINATE GEOMETRY Find the Write an equation using the sine ratio. measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio.

50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – If 5), and D(–1, 2) Use a calculator. SOLUTION:

Use the Distance Formula to find the length of each side. Solve each equation. 2 52. x + 13x = –36

SOLUTION:

Since , triangle BCD is a right Therefore, the solution is –9, –4. triangle. So, CD is the length of the hypotenuse and BC is the 2 length of the leg adjacent to 53. x – 6x = –9 Write an equation using the cosine ratio. SOLUTION:

Use a calculator. Therefore, the solution is 3.

54. 3x2 + 15x = 0 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: SOLUTION: Use the Distance Formula to find the length of each side. Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio. Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 If SOLUTION: Use a calculator.

Solve each equation. 2 Therefore, the solution is –6. 52. x + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

Find each measure. Assume that segments that

appear to be tangent are tangent.

1. 4.

SOLUTION: SOLUTION:

5. 2.

SOLUTION: SOLUTION:

So, the measure of arc TS is 144.

3. So, the measure of arc QTS is 248.

SOLUTION: SOLUTION:

So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several

4. barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes

5. with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION: Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. So, the measure of arc QTS is 248. 8. 6.

SOLUTION: SOLUTION:

So, the measure of arc LP is 150. 9. 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

SOLUTION: 10. Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15. ∠JMK and ∠HMJ form a linear pair.

Find each measure. Assume that segments that appear to be tangent are tangent. 8. 11.

SOLUTION: SOLUTION:

Therefore, the measure of arc RQ is 28. 9. 12.

SOLUTION: SOLUTION:

10. 13.

SOLUTION: SOLUTION:

JMK and HMJ form a linear pair. ∠ ∠ So, the measure of arc PM is 144.

14.

11.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints. SOLUTION:

11-6 Secants, Tangents, and Angle Measures Therefore, the measure of arc RQ is 28. 12. 15.

SOLUTION: SOLUTION:

By Theorem 10.13, .

13.

16.

SOLUTION:

So, the measure of arc PM is 144. SOLUTION:

14. By Theorem 11.13, .

Solve for .

We know that . Substitute.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints. 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

15.

eSolutions Manual - Powered by Cognero Page 3 a. b.

SOLUTION: SOLUTION: a. By Theorem 11.13, . By Theorem 10.13, . Substitute.

Simplify.

b. By Theorem 11.14,

. 16. Substitute.

Simplify.

SOLUTION: CCSS STRUCTURE Find each measure.

By Theorem 11.13, . 18.

Solve for .

We know that . Substitute.

SOLUTION:

By Theorem 11.14, . 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , Substitute. find each measure.

Simplify.

a. 19. b. SOLUTION:

a. By Theorem 11.13, . Substitute.

SOLUTION: Simplify. By Theorem 11.14, . b. By Theorem 11.14, Substitute. . Substitute. Simplify.

Simplify.

CCSS STRUCTURE Find each measure. 20. m(arc JNM) 18.

SOLUTION:

By Theorem 11.14, . Substitute. SOLUTION:

Simplify. So, the measure of arc JNM is 205.

21.

19.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute.

By Theorem 11.14, . Simplify. Substitute.

Simplify. 22.

20. m(arc JNM) SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION: 23.

So, the measure of arc JNM is 205. SOLUTION:

21. By Theorem 11.14, . Substitute.

Solve for .

SOLUTION: 24. JEWELRY In the circular necklace shown, A and B By Theorem 11.14, . are tangent points. If x = 260, what is y? Substitute.

Simplify.

22.

SOLUTION:

By Theorem 11.14, .

SOLUTION: Substitute.

By Theorem 11.14, . Simplify. Substitute.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

23.

SOLUTION: SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem By Theorem 11.14, . 11.14 to find the value of x. Substitute.

Solve for . Therefore, the measure of the planet’s arc that is visible to the satellite is 168. ALGEBRA Find the value of x.

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

26. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

27. 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible SOLUTION: to the satellite. By Theorem 11.14, . Solve for .

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

28. Therefore, the measure of the planet’s arc that is visible to the satellite is 168. SOLUTION:

ALGEBRA Find the value of x. By Theorem 11.14, . Solve for .

26. SOLUTION:

By Theorem 11.14, . Solve for .

27. SOLUTION:

By Theorem 11.14, . Solve for . SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

28. b. Let x be the measure of the camera’s viewing SOLUTION: angle.

By Theorem 11.14, . By Theorem 11.14, . Solve for . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

29. PHOTOGRAPHY A photographer frames a Prove: carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used? SOLUTION: Statements (Reasons) 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem) SOLUTION: 4. (Substitution) a. Let x be the measure of the carousel that appears in the shot. 5. (Subtraction Prop.)

By Theorem 11.14, . 6. (Distributive Prop.) Solve for .

31. Case 2 Given: tangent and secant

Prove: b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . Solve for . SOLUTION: Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. meas. of an inscribed the measure of its 30. Case 1 Given: secants and intercepted arc.) 3. (Exterior Prove: Theorem)

4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.) SOLUTION: Statements (Reasons) 32. Case 3 1. and are secants to the circle. (Given) Given: tangent and 2. , (The Prove: measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution) SOLUTION: 5. (Subtraction Prop.) Statements (Reasons) 1. and are tangents to the circle. (Given) 6. (Distributive Prop.) 2. (The meas. 31. Case 2 of an secant-tangent the measure of its Given: tangent and secant intercepted arc.) Prove: 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.)

SOLUTION: 6. (Distributive Prop.) Statements (Reasons) 1. is a tangent to the circle and is a secant to 33. PROOF Write a paragraph proof of Theorem 11.13.

the circle. (Given) 2. , ( The

meas. of an inscribed the measure of its

intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) a. Given: is a tangent of . is a secant of .

6. (Distributive Prop.) ∠CAE is acute. Prove: m∠CAE = m(arc CA) 32. Case 3 b. Prove that if ∠CAB is obtuse, m∠CAB = m Given: tangent and (arc CDA)

b. Proof: Using the Angle and Arc Addition 5. (Subtraction Prop.) Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since 6. (Distributive Prop.) and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a 33. PROOF Write a paragraph proof of Theorem 11.13. measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since

∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m CAB = m(arc ∠ CF) + m(arc CDA) – m(arc CF). Then by

subtraction, m∠CAB = m(arc CDA).

a. Given: is a tangent of . is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA)

b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA) 34. WALLPAPER The design shown is an example of SOLUTION: optical wallpaper. is a diameter of If m∠A a. Proof: By Theorem 11.10, . So, ∠FAE = 26 and = 67, what is is a right ∠ with measure of 90 and arc FCA is a Refer to the image on page 766. semicircle with measure of 180. Since ∠CAE is SOLUTION: acute, C is in the interior of ∠FAE, so by the Angle First, find the measure of arc BD. and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + Since is a diameter, arc BDE is a semicircle and has a measure of 180. (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution.

m∠FAC = m(arc FC) since ∠FAC is inscribed, so 35. MULTIPLE REPRESENTATIONS In this substitution yields m(arc FC) + m CAE = m ∠ problem, you will explore the relationship between (arc FC) + m(arc CA). By Subtraction Prop., Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then m∠CAE = m(arc CA). draw three successive figures in which the position of point D moves closer to point C, but points A, B, b. Proof: Using the Angle and Arc Addition and C remain fixed. Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since b. TABULAR Estimate the measure of for and is a diameter, ∠FAB is a right angle with a each successive circle, recording the measures of measure of 90 and arc FDA is a semicircle with a and in a table. Then calculate and record measure of 180. By substitution, m∠CAB = m∠CAF the value of x for each circle. + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by c. VERBAL Describe the relationship between substitution, m∠CAB = m(arc CF) + 90. Using the and the value of x as approaches zero. Division and Subtraction Properties on the Arc What type of angle does ∠AEB become when Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m∠CAB = m(arc d. ANALYTICAL Write an algebraic proof to CF) + m(arc CDA) – m(arc CF). Then by show the relationship between Theorems 11.12 and 11.6 described in part c. subtraction, m∠CAB = m(arc CDA).

SOLUTION: 34. WALLPAPER The design shown is an example of a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. optical wallpaper. is a diameter of If m∠A Then draw chords and . = 26 and = 67, what is Refer to the image on page 766. SOLUTION:

First, find the measure of arc BD. b.

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

c. As the measure of gets closer to 0, the 35. MULTIPLE REPRESENTATIONS In this measure of x approaches half of problem, you will explore the relationship between Theorems 11.12 and 11.6. becomes an inscribed angle. a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position d. of point D moves closer to point C, but points A, B, and C remain fixed. 36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a b. TABULAR Estimate the measure of for tangent that intersect outside a circle. each successive circle, recording the measures of SOLUTION: and in a table. Then calculate and record Find the difference of the two intercepted arcs and the value of x for each circle. divide by 2.

c. VERBAL Describe the relationship between 37. CHALLENGE The circles below are concentric.

and the value of x as approaches zero. What is x? What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c. SOLUTION:

SOLUTION: a. Redraw the circle with points A, B, and C in the Use the arcs intercepted on the smaller circle to find same place but place point D closer to C each time. m∠A. Then draw chords and .

b. Use the arcs intercepted on the larger circle and m∠A to find the value of x.

38. REASONS Isosceles is inscribed in c. As the measure of gets closer to 0, the What can you conclude about and measure of x approaches half of Explain. becomes an inscribed angle.

36. WRITING IN MATH Explain how to find the

measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: SOLUTION: Sample answer: m∠BAC = m∠BCA Find the difference of the two intercepted arcs and because the triangle is isosceles. Since ∠BAC and divide by 2. ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. 37. CHALLENGE The circles below are concentric. So, m(arc AB) = m(arc BC). What is x?

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain. SOLUTION:

SOLUTION: a. ; for all values except when Use the arcs intercepted on the smaller circle to find at G, then . m∠A. b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer. Use the arcs intercepted on the larger circle and m∠A to find the value of x. 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. SOLUTION: 38. REASONS Isosceles is inscribed in Sample answer: What can you conclude about and Explain.

By Theorem 11.13, So, 50° =

SOLUTION: Therefore, x (minor arc) = 130, Sample answer: m∠BAC = m∠BCA and y (major arc) = 360° – 130° or 230°. because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m 41. WRITING IN MATH A circle is inscribed within (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. If m∠P = 50 and m∠Q = 60, describe how So, m(arc AB) = m(arc BC). to find the measures of the three minor arcs formed by the points of tangency. SOLUTION: 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

Sample answer: Use Theorem 11.14 to find each minor arc.

SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70. x degrees. Hence solving leads to the answer.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to Therefore, the measures of the three minor arcs are measure the angle that is formed. Find the measures 130, 120, and 110. of the minor and major arcs formed. Explain your reasoning. 42. What is the value of x if and

SOLUTION: Sample answer:

A 23° B 31° By Theorem 11.13, So, 50° = C 64° D 128° Therefore, x (minor arc) = 130, SOLUTION: and y (major arc) = 360° – 130° or 230°. By Theorem 11.14, . 41. WRITING IN MATH A circle is inscribed within Substitute. If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. Simplify. SOLUTION:

Sample answer: Use Theorem 11.14 to find each minor arc. So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) The sum of the angles of a triangle is 180, so m∠R = J (1, 5) 180 – (50 + 60) or 70. SOLUTION: Here, C is the midpoint of . Use the midpoint formula,

Therefore, the measures of the three minor arcs are , to find the coordinates of 130, 120, and 110. C.

42. What is the value of x if and

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC? A 23° B 31° C 64° D 128° SOLUTION: By Theorem 11.14, . SOLUTION:

Substitute. Since .

We know that vertical angles are congruent. So, . Simplify. Since ,

. We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute.

Since , .

So, the correct choice is C. 45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on regions? circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) 2 H (5, –3) A 64π units J (1, 5) B 32π units2 2 SOLUTION: C 12π units 2 Here, C is the midpoint of . D 8π units 2 Use the midpoint formula, D 2π units , to find the coordinates of SOLUTION: C. First, use the circumference to find the radius of the circle.

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and The shaded regions of the circle comprise half of the what is m∠BAC? circle, so its area is half the area of the circle.

SOLUTION: The area of the shaded regions is 32π. Since . So, the correct choice is B. We know that vertical angles are congruent. So, . Find x. Assume that segments that appear to be tangent are tangent. Since , . We know that the sum of the measures of all interior angles of a triangle is 180. 46. Substitute. SOLUTION:

By Theorem 11.10, . So, is a right triangle. , . Since Use the Pythagorean Theorem.

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded Substitute. regions?

2 A 64π units B 32π units2 2 C 12π units D 8π units2 2 47. D 2π units SOLUTION: SOLUTION: If two segments from the same exterior point are First, use the circumference to find the radius of the tangent to a circle, then they are congruent. circle.

The shaded regions of the circle comprise half of the 48. circle, so its area is half the area of the circle. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x.

The area of the shaded regions is 32π. So, the correct choice is B.

Find x. Assume that segments that appear to be 49. PROOF Write a two-column proof. tangent are tangent. Given: is a semicircle;

Prove:

46. SOLUTION: By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem. SOLUTION:

Given: is a semicircle. Substitute.

Prove:

Proof: 47. Statements (Reasons)

SOLUTION: 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle If two segments from the same exterior point are tangent to a circle, then they are congruent. intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.)

7. (Def. of 48. COORDINATE GEOMETRY Find the SOLUTION: measure of each angle to the nearest tenth of a Use Theorem 11.10 and the Pythagorean Theorem. degree by using the Distance Formula and an inverse trigonometric ratio. Solve for x. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each side.

49. PROOF Write a two-column proof. Given: is a semicircle;

Prove:

Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio. SOLUTION:

Given: is a semicircle.

If Prove: Use a calculator.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2)

SOLUTION: Proof: Use the Distance Formula to find the length of each Statements (Reasons) side.

1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.) In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite 7. (Def. of Write an equation using the sine ratio. COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an

inverse trigonometric ratio. If 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) Use a calculator.

SOLUTION: Use the Distance Formula to find the length of each Solve each equation. side. 52. x2 + 13x = –36

SOLUTION:

Therefore, the solution is –9, –4. Since , triangle BCD is a right triangle. 2 So, CD is the length of the hypotenuse and BC is the 53. x – 6x = –9 length of the leg adjacent to SOLUTION: Write an equation using the cosine ratio.

Use a calculator. Therefore, the solution is 3.

54. 3x2 + 15x = 0 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – SOLUTION: 2), and Z(7, –2) SOLUTION: Use the Distance Formula to find the length of each side. Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio. Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0

If SOLUTION:

Use a calculator.

Solve each equation. Therefore, the solution is –6. 52. x2 + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

5. So, the measure of arc TS is 144.

SOLUTION:

SOLUTION: So, the measure of arc QTS is 248.

SOLUTION:

SOLUTION: So, the measure of arc LP is 150.

SOLUTION: SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc So, the measure of arc QTS is 248. with the same endpoints, so its measure is 360 – 165 or 195. 6.

Therefore, the measure of the angle the ramp makes with the ground is 15.

SOLUTION: Find each measure. Assume that segments that appear to be tangent are tangent. 8.

So, the measure of arc LP is 150.

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 SOLUTION: or 195.

Therefore, the measure of the angle the ramp makes 10. with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 8.

SOLUTION:

SOLUTION: ∠JMK and ∠HMJ form a linear pair.

11.

SOLUTION:

Therefore, the measure of arc RQ is 28.

12.

10.

SOLUTION:

∠JMK and ∠HMJ form a linear pair. 13.

11. SOLUTION:

So, the measure of arc PM is 144. SOLUTION: 14.

Therefore, the measure of arc RQ is 28.

12. SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

15.

13.

SOLUTION:

SOLUTION: By Theorem 10.13, .

So, the measure of arc PM is 144.

14. 16.

SOLUTION:

Arc BD and arc BCD are a minor and major arc that share the same endpoints. SOLUTION:

By Theorem 11.13, .

Solve for .

We know that . 15. Substitute.

17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

SOLUTION:

By Theorem 10.13, .

a. b. 16. SOLUTION:

a. By Theorem 11.13, . Substitute.

Simplify.

SOLUTION: b. By Theorem 11.14,

By Theorem 11.13, . .

Solve for . Substitute.

We know that . Simplify. Substitute.

11-6 Secants, Tangents, and Angle Measures

17. SPORTS The multi-sport field shown includes a CCSS STRUCTURE Find each measure.

softball field and a soccer field. If , 18. find each measure.

SOLUTION:

By Theorem 11.14, .

a. Substitute. b.

SOLUTION: Simplify.

a. By Theorem 11.13, . Substitute.

Simplify.

b. By Theorem 11.14, 19.

. Substitute.

Simplify.

SOLUTION:

By Theorem 11.14, . CCSS STRUCTURE Find each measure. 18. Substitute.

Simplify.

SOLUTION:

By Theorem 11.14, . 20. m(arc JNM) Substitute.

eSolutionsSimplify.Manual - Powered by Cognero Page 4

19. SOLUTION:

So, the measure of arc JNM is 205.

21. SOLUTION:

By Theorem 11.14, . Substitute.

Simplify. SOLUTION:

By Theorem 11.14, . Substitute.

m(arc JNM) 20. Simplify.

22.

SOLUTION: SOLUTION: By Theorem 11.14, . Substitute.

So, the measure of arc JNM is 205. Simplify.

21.

23.

SOLUTION: SOLUTION: By Theorem 11.14, . By Theorem 11.14, . Substitute. Substitute.

Simplify. Solve for .

22. 24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify. SOLUTION:

By Theorem 11.14, . Substitute. 23.

Simplify.

SOLUTION: 25. SPACE A satellite orbits above Earth’s equator. By Theorem 11.14, . Find x, the measure of the planet’s arc, that is visible to the satellite. Substitute.

Solve for .

SOLUTION: 24. JEWELRY In the circular necklace shown, A and B The measure of the visible arc is x and the measure are tangent points. If x = 260, what is y? of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x.

SOLUTION: 26. By Theorem 11.14, . SOLUTION: Substitute. By Theorem 11.14, .

Simplify. Solve for .

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

SOLUTION: The measure of the visible arc is x and the measure 27. of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x. SOLUTION: By Theorem 11.14, . Solve for .

Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x.

26. SOLUTION:

By Theorem 11.14, . 28. Solve for . SOLUTION:

By Theorem 11.14, . Solve for .

29. PHOTOGRAPHY A photographer frames a 27. carousel in his camera shot as shown so that the lines

SOLUTION: of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the By Theorem 11.14, . arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in Solve for . the photograph, what viewing angle should be used?

28. SOLUTION: SOLUTION: a. Let x be the measure of the carousel that appears in the shot. By Theorem 11.14, . By Theorem 11.14, . Solve for . Solve for .

b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . . 29. PHOTOGRAPHY A photographer frames a Solve for carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used? CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove:

SOLUTION: SOLUTION: a. Let x be the measure of the carousel that appears Statements (Reasons) in the shot. 1. and are secants to the circle. (Given)

By Theorem 11.14, . 2. , (The Solve for . measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution) b. Let x be the measure of the camera’s viewing angle. 5. (Subtraction Prop.) By Theorem 11.14, .

Solve for . 6. (Distributive Prop.)

31. Case 2 Given: tangent and secant

Prove: CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove: SOLUTION: Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The SOLUTION: Statements (Reasons) meas. of an inscribed the measure of its 1. and are secants to the circle. (Given) intercepted arc.) 2. , (The 3. (Exterior Theorem)

measure of an inscribed the measure of its 4. (Substitution) intercepted arc.) 3. (Exterior 5. (Subtraction Prop.) Theorem) 6. (Distributive Prop.) 4. (Substitution) 32. Case 3 5. (Subtraction Prop.) Given: tangent and

6. (Distributive Prop.) Prove:

31. Case 2 Given: tangent and secant

Prove:

SOLUTION: Statements (Reasons) 1. and are tangents to the circle. (Given)

2. (The meas. SOLUTION: Statements (Reasons) of an secant-tangent the measure of its 1. is a tangent to the circle and is a secant to the circle. (Given) intercepted arc.) 3. (Exterior 2. , ( The Theorem)

meas. of an inscribed the measure of its 4. (Substitution) intercepted arc.) 5. (Subtraction Prop.) 3. (Exterior Theorem) 6. (Distributive Prop.) 4. (Substitution) 33. PROOF Write a paragraph proof of Theorem 11.13. 5. (Subtraction Prop.)

6. (Distributive Prop.)

32. Case 3 Given: tangent and

Prove:

a. Given: is a tangent of . is a secant of . ∠CAE is acute.

Prove: m∠CAE = m(arc CA) SOLUTION: b. Prove that if ∠CAB is obtuse, m∠CAB = m Statements (Reasons) (arc CDA) 1. and are tangents to the circle. (Given)

2. (The meas. SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE of an secant-tangent the measure of its is a right ∠ with measure of 90 and arc FCA is a intercepted arc.) semicircle with measure of 180. Since ∠CAE is 3. (Exterior acute, C is in the interior of ∠FAE, so by the Angle Theorem) and Arc Addition Postulates, m∠FAE = m∠FAC + m CAE and m(arc FCA) = m(arc FC) + m(arc CA). 4. (Substitution) ∠ By substitution, 90 = m∠FAC + m∠CAE and 180 =

5. (Subtraction Prop.) m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE

6. (Distributive Prop.) = m(arc FC) + m(arc CA) by substitution. m∠FAC = m(arc FC) since ∠FAC is inscribed, so 33. PROOF Write a paragraph proof of Theorem 11.13. substitution yields m(arc FC) + m CAE = m ∠ (arc FC) + m(arc CA). By Subtraction Prop.,

m∠CAE = m(arc CA).

b. Proof: Using the Angle and Arc Addition Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m CAB = m CAF a. Given: is a tangent of . ∠ ∠ + 90 and m(arc CDA) = m(arc CF) + 180. Since is a secant of . ∠CAF is inscribed, m∠CAF = m(arc CF) and by ∠CAE is acute. Prove: m∠CAE = m(arc CA) substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc b. Prove that if ∠CAB is obtuse, m∠CAB = m Addition equation yields m(arc CDA) – m(arc (arc CDA) CF) = 90. By substituting for 90, m∠CAB = m(arc SOLUTION: CF) + m(arc CDA) – m(arc CF). Then by a. Proof: By Theorem 11.10, . So, ∠FAE subtraction, m∠CAB = m(arc CDA). is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + 34. WALLPAPER The design shown is an example of (arc CA) by Division Prop., and m∠FAC + m∠CAE optical wallpaper. is a diameter of If m A = m(arc FC) + m(arc CA) by substitution. ∠ = 26 and = 67, what is m FAC = m(arc FC) since FAC is inscribed, so ∠ ∠ Refer to the image on page 766. substitution yields m(arc FC) + m∠CAE = m SOLUTION: (arc FC) + m(arc CA). By Subtraction Prop., First, find the measure of arc BD. m∠CAE = m(arc CA).

b. Proof: Using the Angle and Arc Addition Since is a diameter, arc BDE is a semicircle and Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar has a measure of 180. CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF 35. MULTIPLE REPRESENTATIONS In this + 90 and m(arc CDA) = m(arc CF) + 180. Since problem, you will explore the relationship between

∠CAF is inscribed, m∠CAF = m(arc CF) and by Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then substitution, m∠CAB = m(arc CF) + 90. Using the draw three successive figures in which the position Division and Subtraction Properties on the Arc of point D moves closer to point C, but points A, B,

Addition equation yields m(arc CDA) – m(arc and C remain fixed.

CF) = 90. By substituting for 90, m∠CAB = m(arc b. TABULAR Estimate the measure of for CF) + m(arc CDA) – m(arc CF). Then by each successive circle, recording the measures of subtraction, m∠CAB = m(arc CDA). and in a table. Then calculate and record the value of x for each circle.

c. VERBAL Describe the relationship between

and the value of x as approaches zero. What type of angle does ∠AEB become when

34. WALLPAPER The design shown is an example of d. ANALYTICAL Write an algebraic proof to optical wallpaper. is a diameter of If m∠A show the relationship between Theorems 11.12 and 11.6 described in part c. = 26 and = 67, what is Refer to the image on page 766. SOLUTION: First, find the measure of arc BD.

Since is a diameter, arc BDE is a semicircle and SOLUTION: has a measure of 180. a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and .

35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then b. draw three successive figures in which the position of point D moves closer to point C, but points A, B, and C remain fixed.

b. TABULAR Estimate the measure of for each successive circle, recording the measures of and in a table. Then calculate and record c. As the measure of gets closer to 0, the the value of x for each circle. measure of x approaches half of becomes an inscribed angle. c. VERBAL Describe the relationship between and the value of x as approaches zero. d. What type of angle does ∠AEB become when

36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a d. ANALYTICAL Write an algebraic proof to tangent that intersect outside a circle. show the relationship between Theorems 11.12 and SOLUTION: 11.6 described in part c. Find the difference of the two intercepted arcs and divide by 2.

37. CHALLENGE The circles below are concentric. What is x?

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and . SOLUTION:

Use the arcs intercepted on the smaller circle to find m∠A.

c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle. Use the arcs intercepted on the larger circle and d. m∠A to find the value of x.

36. WRITING IN MATH Explain how to find the

measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: 38. REASONS Isosceles is inscribed in Find the difference of the two intercepted arcs and What can you conclude about and divide by 2. Explain. 37. CHALLENGE The circles below are concentric. What is x?

SOLUTION: Sample answer: m∠BAC = m∠BCA

because the triangle is isosceles. Since ∠BAC and SOLUTION: ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC).

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. Use the arcs intercepted on the smaller circle to find a. Describe the range of possible values for m∠G. m∠A. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

Use the arcs intercepted on the larger circle and m∠A to find the value of x. SOLUTION: a. ; for all values except when

at G, then . b. Because a diameter is 38. REASONS Isosceles is inscribed in involved the intercepted arcs measure (180 – x) and What can you conclude about and x degrees. Hence solving leads to Explain. the answer.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. SOLUTION: SOLUTION: Sample answer: m∠BAC = m∠BCA Sample answer: because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC).

39. CCSS ARGUMENTS In the figure, is a By Theorem 11.13, So, 50° = diameter and is a tangent. a. Describe the range of possible values for m∠G. Therefore, x (minor arc) = 130, Explain.

b. If m∠G = 34, find the measures of minor arcs HJ and y (major arc) = 360° – 130° or 230°. and KH. Explain. 41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. SOLUTION: SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and

x degrees. Hence solving leads to Sample answer: Use Theorem 11.14 to find each the answer. minor arc.

Therefore, the measures of the three minor arcs are 130, 120, and 110. By Theorem 11.13, So, 50° = 42. What is the value of x if and Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°.

41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. A 23° SOLUTION: B 31° C 64° D 128° SOLUTION:

By Theorem 11.14, .

Substitute. Sample answer: Use Theorem 11.14 to find each minor arc. Simplify.

The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70.

So, the correct choice is C.

Therefore, the measures of the three minor arcs are 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on 130, 120, and 110. circle C, and is a diameter. What are the coordinates of C? 42. What is the value of x if and F (2, 10) G (10, –6) H (5, –3) J (1, 5) SOLUTION: Here, C is the midpoint of . Use the midpoint formula, A 23° , to find the coordinates of B 31° C 64° C. D 128°

SOLUTION:

By Theorem 11.14, . So, the correct choice is J. Substitute. 44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC? Simplify.

SOLUTION:

Since . We know that vertical angles are congruent. So, the correct choice is C. So, . Since , 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on . circle C, and is a diameter. What are the We know that the sum of the measures of all interior coordinates of C? angles of a triangle is 180. F (2, 10) G (10, –6) Substitute. H (5, –3) J (1, 5) Since , . SOLUTION: Here, C is the midpoint of . 45. SAT/ACT If the circumference of the circle below Use the midpoint formula, is 16π units, what is the total area of the shaded regions? , to find the coordinates of C.

2 A 64π units B 32π units2 So, the correct choice is J. 2 C 12π units 44. GRIDDED RESPONSE If m∠AED = 95 and D 8π units2 2 what is m∠BAC? D 2π units SOLUTION: First, use the circumference to find the radius of the circle.

SOLUTION:

Since .

We know that vertical angles are congruent. The shaded regions of the circle comprise half of the So, . circle, so its area is half the area of the circle. Since , . We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute.

The area of the shaded regions is 32π. Since , . So, the correct choice is B.

45. SAT/ACT If the circumference of the circle below Find x. Assume that segments that appear to be is 16π units, what is the total area of the shaded tangent are tangent. regions?

46. SOLUTION: 2 A 64π units By Theorem 11.10, . So, is a right 2 B 32π units triangle. 2 C 12π units Use the Pythagorean Theorem.

D 8π units2 2 Substitute. D 2π units SOLUTION: First, use the circumference to find the radius of the circle.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. 47. SOLUTION: If two segments from the same exterior point are tangent to a circle, then they are congruent.

The area of the shaded regions is 32π. So, the correct choice is B.

Find x. Assume that segments that appear to be 48. tangent are tangent. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x. 46. SOLUTION: By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem. 49. PROOF Write a two-column proof. Substitute. Given: is a semicircle;

Prove:

SOLUTION: Given: is a semicircle. 47. SOLUTION: Prove: If two segments from the same exterior point are tangent to a circle, then they are congruent.

Proof: 48. Statements (Reasons) SOLUTION: 1. MHT is a semicircle; (Given) Use Theorem 11.10 and the Pythagorean Theorem. 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) Solve for x. 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.)

7. (Def. of

49. PROOF Write a two-column proof. COORDINATE GEOMETRY Find the Given: is a semicircle; measure of each angle to the nearest tenth of a degree by using the Distance Formula and an Prove: inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each side. SOLUTION:

Given: is a semicircle.

Prove:

Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

Proof: Statements (Reasons)

1. MHT is a semicircle; (Given) If 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) Use a calculator. 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 6. (AA Sim.) 2), and Z(7, –2)

7. (Def. of SOLUTION: Use the Distance Formula to find the length of each

COORDINATE GEOMETRY Find the side. measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each In right triangle XYZ, ZX is the length of the side. hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio.

Use a calculator. Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the Solve each equation. length of the leg adjacent to 52. x2 + 13x = –36 Write an equation using the cosine ratio. SOLUTION:

Use a calculator. Therefore, the solution is –9, –4.

2 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 53. x – 6x = –9 2), and Z(7, –2) SOLUTION: SOLUTION: Use the Distance Formula to find the length of each side.

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION: In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio. Therefore, the solution is –5, 0.

2 If 55. 28 = x + 3x SOLUTION: Use a calculator.

Solve each equation. 52. x2 + 13x = –36 SOLUTION: Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 Therefore, the solution is –6. SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

Find each measure. Assume that segments that appear to be tangent are tangent. SOLUTION: 1.

4. SOLUTION:

SOLUTION:

So, the measure of arc TS is 144.

3. SOLUTION:

So, the measure of arc QTS is 248.

6. SOLUTION:

SOLUTION: 4.

So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several SOLUTION: barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc SOLUTION: with the same endpoints, so its measure is 360 – 165 or 195.

So, the measure of arc QTS is 248.

6. Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 8.

SOLUTION:

So, the measure of arc LP is 150. SOLUTION:

7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION: SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195. 10.

Therefore, the measure of the angle the ramp makes with the ground is 15. SOLUTION: Find each measure. Assume that segments that appear to be tangent are tangent. 8.

∠JMK and ∠HMJ form a linear pair.

SOLUTION: 11.

9. SOLUTION:

Therefore, the measure of arc RQ is 28.

SOLUTION: 12.

10. SOLUTION:

SOLUTION: 13.

JMK and HMJ form a linear pair. ∠ ∠ SOLUTION:

11.

So, the measure of arc PM is 144.

14.

SOLUTION:

SOLUTION: Therefore, the measure of arc RQ is 28. Arc BD and arc BCD are a minor and major arc that share the same endpoints.

12.

15.

SOLUTION:

13. SOLUTION:

By Theorem 10.13, .

SOLUTION:

16.

So, the measure of arc PM is 144.

14.

SOLUTION:

By Theorem 11.13, . SOLUTION: Arc BD and arc BCD are a minor and major arc that Solve for . share the same endpoints. We know that . Substitute.

17. SPORTS The multi-sport field shown includes a

15. softball field and a soccer field. If , find each measure.

SOLUTION:

By Theorem 10.13, . a.

b. SOLUTION:

a. By Theorem 11.13, . Substitute. 16.

Simplify.

b. By Theorem 11.14,

. Substitute.

SOLUTION:

By Theorem 11.13, . Simplify.

Solve for .

We know that . CCSS STRUCTURE Find each measure. Substitute. 18.

17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , find each measure. SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

a. b. SOLUTION:

a. By Theorem 11.13, . Substitute. 19.

Simplify.

b. By Theorem 11.14,

Substitute. SOLUTION: By Theorem 11.14, . Simplify. Substitute.

Simplify. CCSS STRUCTURE Find each measure. 18.

20. m(arc JNM)

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION:

19. So, the measure of arc JNM is 205.

21.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute. By Theorem 11.14, . Substitute. Simplify.

Simplify.

11-6 Secants, Tangents, and Angle Measures

20. m(arc JNM) 22.

SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

So, the measure of arc JNM is 205. 23.

21.

SOLUTION:

By Theorem 11.14, .

Substitute. SOLUTION: By Theorem 11.14, . Solve for .

Substitute.

Simplify. 24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

22.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute. eSolutions Manual - Powered by Cognero By Theorem 11.14, . Page 5 Substitute. Simplify.

Simplify.

23.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

SOLUTION:

By Theorem 11.14, .

Substitute.

SOLUTION: The measure of the visible arc is x and the measure Solve for . of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y? Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x.

26. SOLUTION:

SOLUTION: By Theorem 11.14, . Solve for . By Theorem 11.14, . Substitute.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

27. SOLUTION:

SOLUTION: By Theorem 11.14, . The measure of the visible arc is x and the measure Solve for . of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x.

28.

26. SOLUTION: SOLUTION: By Theorem 11.14, .

By Theorem 11.14, . Solve for .

Solve for .

By Theorem 11.14, . Solve for .

SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . 28. Solve for . SOLUTION: By Theorem 11.14, . Solve for .

b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . Solve for .

29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. CCSS ARGUMENTS For each case of a. If the camera’s viewing angle is , what is the Theorem 11.14, write a two-column proof. arc measure of the carousel that appears in the shot? 30. Case 1 b. If you want to capture an arc measure of in Given: secants and the photograph, what viewing angle should be used? Prove:

SOLUTION: Statements (Reasons) 1. and are secants to the circle. (Given)

2. , (The SOLUTION: a. Let x be the measure of the carousel that appears measure of an inscribed the measure of its in the shot. intercepted arc.) By Theorem 11.14, . 3. (Exterior Solve for . Theorem)

4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.) b. Let x be the measure of the camera’s viewing angle. 31. Case 2 By Theorem 11.14, . Given: tangent and secant Solve for . Prove:

CCSS ARGUMENTS For each case of SOLUTION: Theorem 11.14, write a two-column proof. 30. Case 1 Statements (Reasons) Given: secants and 1. is a tangent to the circle and is a secant to the circle. (Given) Prove: 2. , ( The

meas. of an inscribed the measure of its intercepted arc.) 3. (Exterior SOLUTION: Theorem)

Statements (Reasons) 4. (Substitution) 1. and are secants to the circle. (Given)

5. (Subtraction Prop.) 2. , (The

6. (Distributive Prop.) measure of an inscribed the measure of its

intercepted arc.) 32. Case 3 3. (Exterior Given: tangent and

Theorem) Prove: 4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

31. Case 2 SOLUTION:

Given: tangent and secant Statements (Reasons)

Prove: 1. and are tangents to the circle. (Given)

2. (The meas.

of an secant-tangent the measure of its

intercepted arc.) 3. (Exterior SOLUTION: Theorem) Statements (Reasons) 1. is a tangent to the circle and is a secant to 4. (Substitution) the circle. (Given) 5. (Subtraction Prop.) 2. , ( The

6. (Distributive Prop.) meas. of an inscribed the measure of its intercepted arc.) 33. PROOF Write a paragraph proof of Theorem 11.13. 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

32. Case 3

Given: tangent and a. Given: is a tangent of .

Prove: is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA) b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA)

SOLUTION: SOLUTION: Statements (Reasons) a. Proof: By Theorem 11.10, . So, ∠FAE is a right with measure of 90 and arc FCA is a 1. and are tangents to the circle. (Given) ∠ semicircle with measure of 180. Since ∠CAE is 2. (The meas. acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + of an secant-tangent the measure of its m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = intercepted arc.) 3. (Exterior m(arc FC) + m(arc CA). So, 90 = m(arc FC) + Theorem) (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution. 4. (Substitution) m FAC = m(arc FC) since FAC is inscribed, so ∠ ∠ 5. (Subtraction Prop.) substitution yields m(arc FC) + m∠CAE = m

6. (Distributive Prop.) (arc FC) + m(arc CA). By Subtraction Prop., m∠CAE = m(arc CA). 33. PROOF Write a paragraph proof of Theorem 11.13. b. Proof: Using the Angle and Arc Addition Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc a. Given: is a tangent of . is a secant of . Addition equation yields m(arc CDA) – m(arc ∠CAE is acute. CF) = 90. By substituting for 90, m∠CAB = m(arc

Prove: m∠CAE = m(arc CA) CF) + m(arc CDA) – m(arc CF). Then by b. Prove that if ∠CAB is obtuse, m∠CAB = m subtraction, m∠CAB = m(arc CDA). (arc CDA)

SOLUTION:

a. Proof: By Theorem 11.10, . So, FAE ∠ is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of FAE, so by the Angle ∠ 34. WALLPAPER The design shown is an example of and Arc Addition Postulates, m∠FAE = m∠FAC + optical wallpaper. is a diameter of If m∠A m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA).

By substitution, 90 = m∠FAC + m∠CAE and 180 = = 26 and = 67, what is m(arc FC) + m(arc CA). So, 90 = m(arc FC) + Refer to the image on page 766. (arc CA) by Division Prop., and m∠FAC + m∠CAE SOLUTION: = m(arc FC) + m(arc CA) by substitution. First, find the measure of arc BD. m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m CAE = m ∠ Since is a diameter, arc BDE is a semicircle and (arc FC) + m(arc CA). By Subtraction Prop., has a measure of 180. m∠CAE = m(arc CA).

b. Proof: Using the Angle and Arc Addition 35. MULTIPLE REPRESENTATIONS In this Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar problem, you will explore the relationship between CDA) = m(arc CF) + m(arc FDA). Since Theorems 11.12 and 11.6. and is a diameter, ∠FAB is a right angle with a a. GEOMETRIC Copy the figure shown. Then measure of 90 and arc FDA is a semicircle with a draw three successive figures in which the position measure of 180. By substitution, m∠CAB = m∠CAF of point D moves closer to point C, but points A, B, + 90 and m(arc CDA) = m(arc CF) + 180. Since and C remain fixed. ∠CAF is inscribed, m∠CAF = m(arc CF) and by b. TABULAR Estimate the measure of for substitution, m∠CAB = m(arc CF) + 90. Using the each successive circle, recording the measures of Division and Subtraction Properties on the Arc and in a table. Then calculate and record Addition equation yields m(arc CDA) – m(arc the value of x for each circle. CF) = 90. By substituting for 90, m∠CAB = m(arc c. VERBAL Describe the relationship between CF) + m(arc CDA) – m(arc CF). Then by and the value of x as approaches zero. subtraction, m∠CAB = m(arc CDA). What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c.

34. WALLPAPER The design shown is an example of optical wallpaper. is a diameter of If m∠A = 26 and = 67, what is Refer to the image on page 766. SOLUTION: SOLUTION: First, find the measure of arc BD. a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and .

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position of point D moves closer to point C, but points A, B, and C remain fixed. c. As the measure of gets closer to 0, the measure of x approaches half of b. TABULAR Estimate the measure of for becomes an inscribed angle. each successive circle, recording the measures of d. and in a table. Then calculate and record the value of x for each circle. 36. WRITING IN MATH Explain how to find the c. VERBAL Describe the relationship between measure of an angle formed by a secant and a tangent that intersect outside a circle. and the value of x as approaches zero. What type of angle does ∠AEB become when SOLUTION: Find the difference of the two intercepted arcs and divide by 2. d. ANALYTICAL Write an algebraic proof to 37. CHALLENGE The circles below are concentric. show the relationship between Theorems 11.12 and What is x? 11.6 described in part c.

SOLUTION: SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and .

Use the arcs intercepted on the smaller circle to find b. m∠A.

Use the arcs intercepted on the larger circle and m A to find the value of x. c. As the measure of gets closer to 0, the ∠ measure of x approaches half of becomes an inscribed angle.

d. 38. REASONS Isosceles is inscribed in 36. WRITING IN MATH Explain how to find the What can you conclude about and measure of an angle formed by a secant and a Explain. tangent that intersect outside a circle. SOLUTION: Find the difference of the two intercepted arcs and divide by 2.

37. CHALLENGE The circles below are concentric. What is x? SOLUTION: Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC).

SOLUTION: 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

Use the arcs intercepted on the smaller circle to find m∠A.

SOLUTION: a. ; for all values except when Use the arcs intercepted on the larger circle and at G, then . m∠A to find the value of x. b. Because a diameter is involved the intercepted arcs measure (180 – x) and

x degrees. Hence solving leads to

the answer. REASONS 38. Isosceles is inscribed in 40. OPEN ENDED Draw a circle and two tangents What can you conclude about and that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures Explain. of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer:

SOLUTION: Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m By Theorem 11.13, So, 50° = (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC). Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°. 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. 41. WRITING IN MATH A circle is inscribed within

a. Describe the range of possible values for m∠G. If m∠P = 50 and m∠Q = 60, describe how Explain. to find the measures of the three minor arcs formed

b. If m∠G = 34, find the measures of minor arcs HJ by the points of tangency. and KH. Explain. SOLUTION:

SOLUTION: a. ; for all values except when Sample answer: Use Theorem 11.14 to find each

at G, then . minor arc. b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer.

OPEN ENDED 40. Draw a circle and two tangents that intersect outside the circle. Use a protractor to The sum of the angles of a triangle is 180, so m∠R = measure the angle that is formed. Find the measures 180 – (50 + 60) or 70. of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer: Therefore, the measures of the three minor arcs are 130, 120, and 110.

42. What is the value of x if and

By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130,

and y (major arc) = 360° – 130° or 230°. A 23° B 31° 41. WRITING IN MATH A circle is inscribed within C 64° If m∠P = 50 and m∠Q = 60, describe how D 128° to find the measures of the three minor arcs formed by the points of tangency. SOLUTION:

SOLUTION: By Theorem 11.14, . Substitute.

Simplify.

Sample answer: Use Theorem 11.14 to find each minor arc.

So, the correct choice is C.

The sum of the angles of a triangle is 180, so m∠R = 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on 180 – (50 + 60) or 70. circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) Therefore, the measures of the three minor arcs are H (5, –3) 130, 120, and 110. J (1, 5) SOLUTION: 42. What is the value of x if and Here, C is the midpoint of . Use the midpoint formula, , to find the coordinates of C.

A 23° B 31° C 64° So, the correct choice is J. D 128° SOLUTION: 44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC? By Theorem 11.14, . Substitute.

Simplify. SOLUTION:

Since . We know that vertical angles are congruent. So, . Since , . We know that the sum of the measures of all interior angles of a triangle is 180. So, the correct choice is C. Substitute. 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? Since , . F (2, 10) G (10, –6) 45. SAT/ACT If the circumference of the circle below H (5, –3) is 16π units, what is the total area of the shaded J (1, 5) regions? SOLUTION: Here, C is the midpoint of . Use the midpoint formula, 2 , to find the coordinates of A 64π units C. B 32π units2 2 C 12π units

D 8π units2 2 D 2π units

So, the correct choice is J. SOLUTION: First, use the circumference to find the radius of the 44. GRIDDED RESPONSE If m∠AED = 95 and circle. what is m∠BAC?

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. SOLUTION:

Since . We know that vertical angles are congruent. So, . Since , . We know that the sum of the measures of all interior The area of the shaded regions is 32π. angles of a triangle is 180. So, the correct choice is B.

Substitute. Find x. Assume that segments that appear to be tangent are tangent.

Since , .

45. SAT/ACT If the circumference of the circle below 46. is 16π units, what is the total area of the shaded regions? SOLUTION: By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem.

2 Substitute. A 64π units

B 32π units2 2 C 12π units D 8π units2 2 D 2π units SOLUTION: First, use the circumference to find the radius of the circle.

47. SOLUTION: If two segments from the same exterior point are tangent to a circle, then they are congruent. The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

48. SOLUTION: The area of the shaded regions is 32π. Use Theorem 11.10 and the Pythagorean Theorem. So, the correct choice is B. Solve for x. Find x. Assume that segments that appear to be

tangent are tangent.

46. 49. PROOF Write a two-column proof. SOLUTION:

By Theorem 11.10, . So, is a right Given: is a semicircle;

triangle. Prove: Use the Pythagorean Theorem.

Substitute.

SOLUTION:

Given: is a semicircle.

Prove:

47. SOLUTION: If two segments from the same exterior point are tangent to a circle, then they are congruent.

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 48. 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) SOLUTION: 5. (Reflexive Prop.) Use Theorem 11.10 and the Pythagorean Theorem. 6. (AA Sim.)

Solve for x. 7. (Def. of

COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 49. PROOF Write a two-column proof. 5), and D(–1, 2)

Given: is a semicircle; SOLUTION:

Prove: Use the Distance Formula to find the length of each side.

SOLUTION:

Given: is a semicircle. Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the Prove: length of the leg adjacent to Write an equation using the cosine ratio.

Use a calculator. Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2. is a right angle. (If an inscribed angle 2), and Z(7, –2) intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) SOLUTION: 4. (All rt. angles are .) Use the Distance Formula to find the length of each 5. (Reflexive Prop.) side. 6. (AA Sim.)

7. (Def. of

COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite 50. C in triangle BCD with vertices B(–1, –5), C(–6, – ∠ 5), and D(–1, 2) Write an equation using the sine ratio. SOLUTION: Use the Distance Formula to find the length of each side. If

Use a calculator.

Solve each equation. 52. x2 + 13x = –36 Since , triangle BCD is a right SOLUTION: triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

Therefore, the solution is –9, –4.

2 If 53. x – 6x = –9 Use a calculator. SOLUTION:

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Therefore, the solution is 3. Use the Distance Formula to find the length of each side. 54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0. In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite 2 55. 28 = x + 3x

Write an equation using the sine ratio. SOLUTION:

Use a calculator. Therefore, the solution is –7, 4.

Solve each equation. 56. x2 + 12x + 36 = 0 2 52. x + 13x = –36 SOLUTION: SOLUTION:

Therefore, the solution is –6.

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION: SOLUTION:

2. 5.

SOLUTION: SOLUTION:

So, the measure of arc TS is 144.

3. So, the measure of arc QTS is 248.

SOLUTION:

So, the measure of arc LP is 150. 4. 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

5. SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

So, the measure of arc QTS is 248. Find each measure. Assume that segments that appear to be tangent are tangent.

6. 8.

SOLUTION: SOLUTION:

So, the measure of arc LP is 150. 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus 9. motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting 10. tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that ∠JMK and ∠HMJ form a linear pair. appear to be tangent are tangent. 8.

11.

SOLUTION:

9. Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

10.

13.

SOLUTION:

∠JMK and ∠HMJ form a linear pair.

So, the measure of arc PM is 144.

14. 11.

SOLUTION: SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

Therefore, the measure of arc RQ is 28.

12. 15.

SOLUTION:

By Theorem 10.13, .

13.

16. SOLUTION:

So, the measure of arc PM is 144.

14. SOLUTION:

By Theorem 11.13, .

Solve for .

We know that . SOLUTION: Substitute. Arc BD and arc BCD are a minor and major arc that share the same endpoints.

17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

15.

a. b. SOLUTION: SOLUTION: By Theorem 10.13, . a. By Theorem 11.13, .

Substitute.

Simplify.

b 16. . By Theorem 11.14,

. Substitute.

Simplify.

SOLUTION:

By Theorem 11.13, . CCSS STRUCTURE Find each measure. 18. Solve for .

We know that . Substitute.

SOLUTION: 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , By Theorem 11.14, . find each measure. Substitute.

Simplify.

a. b. 19. SOLUTION:

a. By Theorem 11.13, . Substitute.

Simplify. SOLUTION: b. By Theorem 11.14, By Theorem 11.14, . . Substitute.

Substitute.

Simplify.

CCSS STRUCTURE Find each measure. 18. 20. m(arc JNM)

SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

So, the measure of arc JNM is 205.

21. 19.

SOLUTION: SOLUTION: By Theorem 11.14, .

By Theorem 11.14, . Substitute.

Substitute. Simplify.

Simplify.

22.

20. m(arc JNM)

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION: 23.

So, the measure of arc JNM is 205.

21. SOLUTION:

By Theorem 11.14, . Substitute.

Solve for .

SOLUTION:

By Theorem 11.14, . 24. JEWELRY In the circular necklace shown, A and B Substitute. are tangent points. If x = 260, what is y?

Simplify.

22.

SOLUTION: SOLUTION: By Theorem 11.14, .

By Theorem 11.14, . Substitute.

Substitute. Simplify.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible 23. to the satellite.

SOLUTION: SOLUTION: By Theorem 11.14, . The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem Substitute. 11.14 to find the value of x.

Solve for .

11-6 Secants, Tangents, and Angle Measures Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

24. JEWELRY In the circular necklace shown, A and B ALGEBRA Find the value of x. are tangent points. If x = 260, what is y?

26. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible 27. to the satellite. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is visible to the satellite is 168. 28. ALGEBRA Find the value of x. SOLUTION:

By Theorem 11.14, . Solve for . 26. eSolutionsSOLUTION: Manual - Powered by Cognero Page 6

By Theorem 11.14, . Solve for .

27. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

28. SOLUTION: b. By Theorem 11.14, . Let x be the measure of the camera’s viewing angle. Solve for . By Theorem 11.14, . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 29. PHOTOGRAPHY A photographer frames a Given: secants and carousel in his camera shot as shown so that the lines

of sight form tangents to the carousel. Prove: a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

SOLUTION: Statements (Reasons) 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) SOLUTION: 3. (Exterior Theorem) a. Let x be the measure of the carousel that appears

in the shot. 4. (Substitution)

By Theorem 11.14, . 5. (Subtraction Prop.) Solve for .

6. (Distributive Prop.)

31. Case 2 Given: tangent and secant

b. Let x be the measure of the camera’s viewing Prove: angle.

By Theorem 11.14, . Solve for .

SOLUTION: Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given) CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 2. , ( The 30. Case 1 Given: secants and meas. of an inscribed the measure of its

Prove: intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) SOLUTION:

Statements (Reasons) 6. (Distributive Prop.) 1. and are secants to the circle. (Given) 2. , (The 32. Case 3 Given: tangent and

measure of an inscribed the measure of its Prove: intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) SOLUTION: 6. (Distributive Prop.) Statements (Reasons) 1. and are tangents to the circle. (Given)

31. Case 2 2. (The meas. Given: tangent and secant of an secant-tangent the measure of its Prove: intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

SOLUTION: 5. (Subtraction Prop.) Statements (Reasons) 1. is a tangent to the circle and is a secant to 6. (Distributive Prop.) the circle. (Given) 33. PROOF Write a paragraph proof of Theorem 11.13. 2. , ( The

meas. of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) a. Given: is a tangent of . 6. (Distributive Prop.) is a secant of . CAE 32. Case 3 ∠ is acute.

Given: tangent and Prove: m∠CAE = m(arc CA) b. Prove that if ∠CAB is obtuse, m∠CAB = m Prove: (arc CDA)

SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is SOLUTION: acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m FAE = m FAC + Statements (Reasons) ∠ ∠ m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). 1. and are tangents to the circle. (Given) By substitution, 90 = m∠FAC + m∠CAE and 180 = 2. (The meas. m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE of an secant-tangent the measure of its = m(arc FC) + m(arc CA) by substitution. intercepted arc.) m∠FAC = m(arc FC) since ∠FAC is inscribed, so 3. (Exterior Theorem) substitution yields m(arc FC) + m∠CAE = m

4. (Substitution) (arc FC) + m(arc CA). By Subtraction Prop.,

m∠CAE = m(arc CA). 5. (Subtraction Prop.) b. Proof: Using the Angle and Arc Addition 6. (Distributive Prop.) Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since 33. PROOF Write a paragraph proof of Theorem 11.13. and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m CAB = m(arc CF) + 90. Using the ∠ Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc CF) = 90. By substituting for 90, m∠CAB = m(arc

CF) + m(arc CDA) – m(arc CF). Then by a. Given: is a tangent of .

is a secant of . subtraction, m∠CAB = m(arc CDA). ∠CAE is acute.

Prove: m∠CAE = m(arc CA)

b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA)

SOLUTION: 34. WALLPAPER The design shown is an example of a. Proof: By Theorem 11.10, . So, ∠FAE optical wallpaper. is a diameter of If m∠A is a right with measure of 90 and arc FCA is a ∠ semicircle with measure of 180. Since ∠CAE is = 26 and = 67, what is acute, C is in the interior of ∠FAE, so by the Angle Refer to the image on page 766. and Arc Addition Postulates, m∠FAE = m∠FAC + SOLUTION: m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). First, find the measure of arc BD. By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) +

(arc CA) by Division Prop., and m∠FAC + m∠CAE Since is a diameter, arc BDE is a semicircle and = m(arc FC) + m(arc CA) by substitution. has a measure of 180. m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m (arc FC) + m(arc CA). By Subtraction Prop., 35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between m∠CAE = m(arc CA). Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then b. Proof: Using the Angle and Arc Addition draw three successive figures in which the position Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar of point D moves closer to point C, but points A, B, CDA) = m(arc CF) + m(arc FDA). Since and C remain fixed.

and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a b. TABULAR Estimate the measure of for measure of 180. By substitution, m∠CAB = m∠CAF each successive circle, recording the measures of + 90 and m(arc CDA) = m(arc CF) + 180. Since and in a table. Then calculate and record ∠CAF is inscribed, m∠CAF = m(arc CF) and by the value of x for each circle.

substitution, m∠CAB = m(arc CF) + 90. Using the c. VERBAL Describe the relationship between Division and Subtraction Properties on the Arc and the value of x as approaches zero. Addition equation yields m(arc CDA) – m(arc What type of angle does ∠AEB become when CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by d. ANALYTICAL Write an algebraic proof to subtraction, m∠CAB = m(arc CDA). show the relationship between Theorems 11.12 and 11.6 described in part c.

34. WALLPAPER The design shown is an example of SOLUTION: optical wallpaper. is a diameter of If m∠A a. Redraw the circle with points A, B, and C in the = 26 and = 67, what is same place but place point D closer to C each time. Refer to the image on page 766. Then draw chords and . SOLUTION: First, find the measure of arc BD.

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between c. As the measure of gets closer to 0, the Theorems 11.12 and 11.6. measure of x approaches half of a. GEOMETRIC Copy the figure shown. Then becomes an inscribed angle. draw three successive figures in which the position of point D moves closer to point C, but points A, B, and C remain fixed. d.

b. TABULAR Estimate the measure of for 36. WRITING IN MATH Explain how to find the each successive circle, recording the measures of measure of an angle formed by a secant and a tangent that intersect outside a circle. and in a table. Then calculate and record the value of x for each circle. SOLUTION: Find the difference of the two intercepted arcs and c. VERBAL Describe the relationship between divide by 2. and the value of x as approaches zero. 37. CHALLENGE The circles below are concentric.

What type of angle does ∠AEB become when What is x?

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c.

SOLUTION:

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and . Use the arcs intercepted on the smaller circle to find m∠A.

Use the arcs intercepted on the larger circle and m∠A to find the value of x.

c. As the measure of gets closer to 0, the measure of x approaches half of 38. REASONS Isosceles is inscribed in becomes an inscribed angle. What can you conclude about and Explain. d.

36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a

tangent that intersect outside a circle. SOLUTION: SOLUTION: Find the difference of the two intercepted arcs and divide by 2. Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and 37. CHALLENGE The circles below are concentric. ∠BCA are inscribed angles, by Theorem 11.6, m What is x? (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC).

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. SOLUTION: b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

Use the arcs intercepted on the smaller circle to find SOLUTION: m∠A. a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and

Use the arcs intercepted on the larger circle and x degrees. Hence solving leads to m∠A to find the value of x. the answer.

SOLUTION: By Theorem 11.13, So, 50° = Sample answer: m∠BAC = m∠BCA Therefore, x (minor arc) = 130, because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m and y (major arc) = 360° – 130° or 230°. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC). 41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. SOLUTION: a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

Sample answer: Use Theorem 11.14 to find each minor arc. SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and

x degrees. Hence solving leads to The sum of the angles of a triangle is 180, so m∠R = the answer. 180 – (50 + 60) or 70.

SOLUTION: 42. What is the value of x if and Sample answer:

A 23° By Theorem 11.13, So, 50° = B 31° C 64° Therefore, x (minor arc) = 130, D 128° and y (major arc) = 360° – 130° or 230°. SOLUTION: 41. WRITING IN MATH A circle is inscribed within By Theorem 11.14, . If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed Substitute. by the points of tangency. SOLUTION: Simplify.

Sample answer: Use Theorem 11.14 to find each minor arc.

So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) The sum of the angles of a triangle is 180, so m∠R = G (10, –6) 180 – (50 + 60) or 70. H (5, –3) J (1, 5) SOLUTION: Here, C is the midpoint of . Therefore, the measures of the three minor arcs are Use the midpoint formula, 130, 120, and 110. , to find the coordinates of 42. What is the value of x if and C.

So, the correct choice is J.

A 23° 44. GRIDDED RESPONSE If m∠AED = 95 and B 31° what is m∠BAC? C 64° D 128° SOLUTION:

By Theorem 11.14, . Substitute. SOLUTION:

Since . Simplify. We know that vertical angles are congruent. So, . Since , . We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute.

So, the correct choice is C. Since , .

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on 45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded circle C, and is a diameter. What are the regions? coordinates of C? F (2, 10) G (10, –6) H (5, –3) J (1, 5) 2 A 64π units SOLUTION: B 32π units2 Here, C is the midpoint of . 2 C 12π units Use the midpoint formula, D 8π units2 , to find the coordinates of 2 D 2π units C. SOLUTION: First, use the circumference to find the radius of the circle.

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC? The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

SOLUTION:

Since . We know that vertical angles are congruent. The area of the shaded regions is 32π. So, . So, the correct choice is B. Since , Find x. Assume that segments that appear to be . tangent are tangent. We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute. 46. SOLUTION: Since , . By Theorem 11.10, . So, is a right triangle. SAT/ACT 45. If the circumference of the circle below Use the Pythagorean Theorem. is 16π units, what is the total area of the shaded regions? Substitute.

2 A 64π units B 32π units2 2 C 12π units D 8π units2 2 D 2π units SOLUTION: 47. First, use the circumference to find the radius of the SOLUTION: circle. If two segments from the same exterior point are tangent to a circle, then they are congruent.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. 48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x.

The area of the shaded regions is 32π. So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent. 49. PROOF Write a two-column proof. Given: is a semicircle;

Prove: 46. SOLUTION: By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem.

Substitute. SOLUTION: Given: is a semicircle.

Prove:

47. Proof: SOLUTION: Statements (Reasons) If two segments from the same exterior point are 1. MHT is a semicircle; (Given) tangent to a circle, then they are congruent. 2. is a right angle. (If an inscribed angle

intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.)

48. 7. (Def. of SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem. COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a Solve for x. degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each side. 49. PROOF Write a two-column proof.

Given: is a semicircle;

Prove:

Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the SOLUTION: length of the leg adjacent to Given: is a semicircle. Write an equation using the cosine ratio.

Prove: If

Use a calculator.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) Proof: Statements (Reasons) SOLUTION: 1. MHT is a semicircle; (Given) Use the Distance Formula to find the length of each side. 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.)

7. (Def. of In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a Write an equation using the sine ratio. degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) If

SOLUTION: Use a calculator. Use the Distance Formula to find the length of each side. Solve each equation. 2 52. x + 13x = –36 SOLUTION:

Since , triangle BCD is a right triangle. Therefore, the solution is –9, –4. So, CD is the length of the hypotenuse and BC is the

length of the leg adjacent to 2 Write an equation using the cosine ratio. 53. x – 6x = –9 SOLUTION:

Use a calculator.

Therefore, the solution is 3.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 54. 3x2 + 15x = 0 2), and Z(7, –2) SOLUTION: SOLUTION: Use the Distance Formula to find the length of each side.

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION: In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio.

Therefore, the solution is –7, 4.

If 56. x2 + 12x + 36 = 0 Use a calculator. SOLUTION:

Solve each equation. 52. x2 + 13x = –36 Therefore, the solution is –6. SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144. SOLUTION: 3.

SOLUTION:

4. So, the measure of arc TS is 144.

SOLUTION:

SOLUTION: 5.

SOLUTION:

SOLUTION: So, the measure of arc QTS is 248.

SOLUTION:

SOLUTION: So, the measure of arc LP is 150. 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground? So, the measure of arc QTS is 248.

SOLUTION: SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 So, the measure of arc LP is 150. or 195.

7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 8.

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc SOLUTION: with the same endpoints, so its measure is 360 – 165 or 195.

Therefore, the measure of the angle the ramp makes 9. with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 8.

SOLUTION:

SOLUTION: 10.

9. SOLUTION:

∠JMK and ∠HMJ form a linear pair. SOLUTION:

11.

10.

SOLUTION:

Therefore, the measure of arc RQ is 28.

12. ∠JMK and ∠HMJ form a linear pair.

11.

SOLUTION:

SOLUTION: 13.

Therefore, the measure of arc RQ is 28.

12. SOLUTION:

So, the measure of arc PM is 144.

SOLUTION: 14.

13. SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

15. So, the measure of arc PM is 144.

14.

SOLUTION:

By Theorem 10.13, . SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

16. 15.

SOLUTION:

SOLUTION: By Theorem 11.13, .

By Theorem 10.13, . Solve for .

We know that . Substitute.

16. 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

SOLUTION:

By Theorem 11.13, .

Solve for . a. b. We know that . SOLUTION: Substitute. a. By Theorem 11.13, .

Substitute.

17. SPORTS The multi-sport field shown includes a Simplify. softball field and a soccer field. If ,

find each measure. b. By Theorem 11.14,

. Substitute.

Simplify.

a. b. CCSS STRUCTURE Find each measure. SOLUTION: 18.

a. By Theorem 11.13, . Substitute.

Simplify.

b. By Theorem 11.14, SOLUTION:

. By Theorem 11.14, . Substitute. Substitute.

Simplify. Simplify.

CCSS STRUCTURE Find each measure. 18. 19.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute. By Theorem 11.14, . Substitute. Simplify.

Simplify.

19.

20. m(arc JNM)

SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

So, the measure of arc JNM is 205.

20. m(arc JNM) 21.

SOLUTION:

By Theorem 11.14, . Substitute. SOLUTION:

Simplify.

So, the measure of arc JNM is 205.

22. 21.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute.

By Theorem 11.14, . Substitute. Simplify.

Simplify.

23.

22.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute. By Theorem 11.14, . Substitute. Solve for .

Simplify.

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

23.

SOLUTION:

By Theorem 11.14, . Substitute. SOLUTION:

By Theorem 11.14, . Solve for . Substitute.

Simplify. 24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

SOLUTION: SOLUTION: The measure of the visible arc is x and the measure By Theorem 11.14, . of the arc that is not visible is 360 – x. Use Theorem Substitute. 11.14 to find the value of x.

Simplify.

Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x. 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

26. SOLUTION:

By Theorem 11.14, .

SOLUTION: Solve for . The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x.

26. 27. SOLUTION: SOLUTION:

By Theorem 11.14, . By Theorem 11.14, . Solve for . Solve for .

28. SOLUTION:

27. By Theorem 11.14, . SOLUTION: Solve for .

By Theorem 11.14, . Solve for .

11-6 Secants, Tangents, and Angle Measures

29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the 28. arc measure of the carousel that appears in the shot? SOLUTION: b. If you want to capture an arc measure of in the photograph, what viewing angle should be used? By Theorem 11.14, . Solve for .

SOLUTION: a. Let x be the measure of the carousel that appears 29. PHOTOGRAPHY A photographer frames a in the shot.

carousel in his camera shot as shown so that the lines By Theorem 11.14, . of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the Solve for . arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . Solve for .

SOLUTION: a. Let x be the measure of the carousel that appears in the shot. CCSS ARGUMENTS For each case of By Theorem 11.14, . Theorem 11.14, write a two-column proof. 30. Case 1 Solve for . Given: secants and

Prove:

eSolutions Manual - Powered by Cognero Page 7 b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . SOLUTION: Statements (Reasons) Solve for . 1. and are secants to the circle. (Given)

2. , (The

measure of an inscribed the measure of its intercepted arc.) CCSS ARGUMENTS For each case of 3. (Exterior Theorem 11.14, write a two-column proof. Theorem) 30. Case 1 4. (Substitution) Given: secants and

Prove: 5. (Subtraction Prop.)

6. (Distributive Prop.)

31. Case 2

Given: tangent and secant SOLUTION: Statements (Reasons) Prove: 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its

intercepted arc.) SOLUTION: 3. (Exterior Statements (Reasons) Theorem) 1. is a tangent to the circle and is a secant to 4. (Substitution) the circle. (Given) 2. , ( The 5. (Subtraction Prop.) meas. of an inscribed the measure of its 6. (Distributive Prop.) intercepted arc.) 3. (Exterior 31. Case 2 Theorem) Given: tangent and secant 4. (Substitution) Prove:

5. (Subtraction Prop.)

6. (Distributive Prop.)

32. Case 3

SOLUTION: Given: tangent and

Statements (Reasons) Prove: 1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The

meas. of an inscribed the measure of its intercepted arc.) 3. (Exterior SOLUTION: Theorem) Statements (Reasons)

4. (Substitution) 1. and are tangents to the circle. (Given) 2. (The meas. 5. (Subtraction Prop.) of an secant-tangent the measure of its 6. (Distributive Prop.) intercepted arc.) 3. (Exterior 32. Case 3 Theorem)

Given: tangent and 4. (Substitution) Prove:

5. (Subtraction Prop.)

6. (Distributive Prop.)

33. PROOF Write a paragraph proof of Theorem 11.13.

SOLUTION: Statements (Reasons) 1. and are tangents to the circle. (Given)

2. (The meas.

of an secant-tangent the measure of its

intercepted arc.)

3. (Exterior

Theorem) a. Given: is a tangent of . is a secant of . 4. (Substitution) ∠CAE is acute. Prove: m CAE = m(arc CA) 5. (Subtraction Prop.) ∠ b. Prove that if CAB is obtuse, m CAB = m ∠ ∠ 6. (Distributive Prop.) (arc CDA)

PROOF 33. Write a paragraph proof of Theorem 11.13. SOLUTION:

a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE a. Given: is a tangent of . = m(arc FC) + m(arc CA) by substitution.

is a secant of . m∠FAC = m(arc FC) since ∠FAC is inscribed, so ∠CAE is acute. substitution yields m(arc FC) + m∠CAE = m Prove: m∠CAE = m(arc CA) (arc FC) + m(arc CA). By Subtraction Prop., b. Prove that if ∠CAB is obtuse, m∠CAB = m

(arc CDA) m∠CAE = m(arc CA).

b. Proof: Using the Angle and Arc Addition SOLUTION: Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar a. Proof: By Theorem 11.10, . So, ∠FAE CDA) = m(arc CF) + m(arc FDA). Since is a right ∠ with measure of 90 and arc FCA is a and is a diameter, ∠FAB is a right angle with a semicircle with measure of 180. Since ∠CAE is measure of 90 and arc FDA is a semicircle with a acute, C is in the interior of ∠FAE, so by the Angle measure of 180. By substitution, m∠CAB = m∠CAF and Arc Addition Postulates, m∠FAE = m∠FAC + + 90 and m(arc CDA) = m(arc CF) + 180. Since m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). ∠CAF is inscribed, m∠CAF = m(arc CF) and by By substitution, 90 = m∠FAC + m∠CAE and 180 = substitution, m∠CAB = m(arc CF) + 90. Using the m(arc FC) + m(arc CA). So, 90 = m(arc FC) + Division and Subtraction Properties on the Arc (arc CA) by Division Prop., and m∠FAC + m∠CAE Addition equation yields m(arc CDA) – m(arc = m(arc FC) + m(arc CA) by substitution. CF) = 90. By substituting for 90, m∠CAB = m(arc m∠FAC = m(arc FC) since ∠FAC is inscribed, so CF) + m(arc CDA) – m(arc CF). Then by substitution yields m(arc FC) + m∠CAE = m subtraction, m∠CAB = m(arc CDA). (arc FC) + m(arc CA). By Subtraction Prop.,

m∠CAE = m(arc CA).

b. Proof: Using the Angle and Arc Addition Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a 34. WALLPAPER The design shown is an example of measure of 180. By substitution, m∠CAB = m∠CAF optical wallpaper. is a diameter of If m∠A + 90 and m(arc CDA) = m(arc CF) + 180. Since = 26 and = 67, what is ∠CAF is inscribed, m∠CAF = m(arc CF) and by Refer to the image on page 766. substitution, m∠CAB = m(arc CF) + 90. Using the SOLUTION: Division and Subtraction Properties on the Arc First, find the measure of arc BD. Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by Since is a diameter, arc BDE is a semicircle and has a measure of 180. subtraction, m∠CAB = m(arc CDA).

35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position 34. WALLPAPER The design shown is an example of of point D moves closer to point C, but points A, B, and C remain fixed. optical wallpaper. is a diameter of If m∠A

= 26 and = 67, what is b. TABULAR Estimate the measure of for Refer to the image on page 766. each successive circle, recording the measures of SOLUTION: and in a table. Then calculate and record First, find the measure of arc BD. the value of x for each circle.

c. VERBAL Describe the relationship between

Since is a diameter, arc BDE is a semicircle and and the value of x as approaches zero. has a measure of 180. What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to 35. MULTIPLE REPRESENTATIONS In this show the relationship between Theorems 11.12 and problem, you will explore the relationship between 11.6 described in part c. Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position of point D moves closer to point C, but points A, B, and C remain fixed.

b. TABULAR Estimate the measure of for SOLUTION: each successive circle, recording the measures of a. Redraw the circle with points A, B, and C in the and in a table. Then calculate and record same place but place point D closer to C each time. the value of x for each circle. Then draw chords and .

c. VERBAL Describe the relationship between and the value of x as approaches zero. What type of angle does ∠AEB become when b.

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c.

c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle.

SOLUTION: d. a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. 36. WRITING IN MATH Explain how to find the Then draw chords and . measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: Find the difference of the two intercepted arcs and b. divide by 2. 37. CHALLENGE The circles below are concentric. What is x?

c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle. SOLUTION:

36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle.

SOLUTION: Use the arcs intercepted on the smaller circle to find Find the difference of the two intercepted arcs and m∠A. divide by 2.

37. CHALLENGE The circles below are concentric. What is x?

Use the arcs intercepted on the larger circle and m∠A to find the value of x.

SOLUTION: 38. REASONS Isosceles is inscribed in What can you conclude about and Explain.

Use the arcs intercepted on the smaller circle to find m∠A.

SOLUTION: Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and Use the arcs intercepted on the larger circle and ∠BCA are inscribed angles, by Theorem 11.6, m m∠A to find the value of x. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC).

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. 38. REASONS Isosceles is inscribed in a. Describe the range of possible values for m∠G. What can you conclude about and Explain. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

SOLUTION: SOLUTION: Sample answer: m∠BAC = m∠BCA a. ; for all values except when

because the triangle is isosceles. Since ∠BAC and at G, then . ∠BCA are inscribed angles, by Theorem 11.6, m b. Because a diameter is (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. involved the intercepted arcs measure (180 – x) and So, m(arc AB) = m(arc BC). x degrees. Hence solving leads to the answer. 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to a. Describe the range of possible values for m∠G. measure the angle that is formed. Find the measures Explain. of the minor and major arcs formed. Explain your b. If m∠G = 34, find the measures of minor arcs HJ reasoning. and KH. Explain. SOLUTION: Sample answer:

SOLUTION: a. ; for all values except when

at G, then . By Theorem 11.13, So, 50° = b. Because a diameter is involved the intercepted arcs measure (180 – x) and Therefore, x (minor arc) = 130, x degrees. Hence solving leads to and y (major arc) = 360° – 130° or 230°. the answer. 41. WRITING IN MATH A circle is inscribed within 40. OPEN ENDED Draw a circle and two tangents If m∠P = 50 and m∠Q = 60, describe how that intersect outside the circle. Use a protractor to to find the measures of the three minor arcs formed measure the angle that is formed. Find the measures by the points of tangency. of the minor and major arcs formed. Explain your SOLUTION: reasoning. SOLUTION: Sample answer:

Sample answer: Use Theorem 11.14 to find each minor arc. By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°.

WRITING IN MATH 41. A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how The sum of the angles of a triangle is 180, so m∠R = to find the measures of the three minor arcs formed 180 – (50 + 60) or 70. by the points of tangency. SOLUTION:

Therefore, the measures of the three minor arcs are 130, 120, and 110.

42. What is the value of x if and

Sample answer: Use Theorem 11.14 to find each minor arc.

A 23° B 31° C 64° D 128° The sum of the angles of a triangle is 180, so m∠R = SOLUTION: 180 – (50 + 60) or 70. By Theorem 11.14, . Substitute.

Therefore, the measures of the three minor arcs are 130, 120, and 110. Simplify.

42. What is the value of x if and

A 23° B 31° So, the correct choice is C. C 64° D 128° 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on SOLUTION: circle C, and is a diameter. What are the coordinates of C? By Theorem 11.14, . F (2, 10) G (10, –6) Substitute. H (5, –3) J (1, 5) Simplify. SOLUTION: Here, C is the midpoint of . Use the midpoint formula, , to find the coordinates of C.

So, the correct choice is C. So, the correct choice is J.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on 44. GRIDDED RESPONSE If m∠AED = 95 and

circle C, and is a diameter. What are the what is m∠BAC? coordinates of C? F (2, 10) G (10, –6) H (5, –3) J (1, 5) SOLUTION: SOLUTION:

Here, C is the midpoint of . Since . Use the midpoint formula, We know that vertical angles are congruent. , to find the coordinates of So, . C. Since ,

. We know that the sum of the measures of all interior angles of a triangle is 180.

So, the correct choice is J. Substitute.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC? Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

SOLUTION:

Since . 2 A 64π units 2 We know that vertical angles are congruent. B 32π units So, . 2 C 12π units Since , 2 . D 8π units 2 We know that the sum of the measures of all interior D 2π units angles of a triangle is 180. SOLUTION: Substitute. First, use the circumference to find the radius of the circle.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions? The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

2 A 64π units B 32π units2 2 C 12π units D 8π units2 The area of the shaded regions is 32π. 2 D 2π units So, the correct choice is B. SOLUTION: Find x. Assume that segments that appear to be First, use the circumference to find the radius of the tangent are tangent. circle.

46. SOLUTION: By Theorem 11.10, . So, is a right The shaded regions of the circle comprise half of the triangle. circle, so its area is half the area of the circle. Use the Pythagorean Theorem.

Substitute.

The area of the shaded regions is 32π. So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent.

47. SOLUTION: 46. If two segments from the same exterior point are SOLUTION: tangent to a circle, then they are congruent.

By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem.

Substitute.

48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x.

47. SOLUTION: 49. PROOF Write a two-column proof. If two segments from the same exterior point are tangent to a circle, then they are congruent. Given: is a semicircle;

Prove:

48. SOLUTION: SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem. Given: is a semicircle.

Solve for x. Prove:

49. PROOF Write a two-column proof. Given: is a semicircle; Proof:

Prove: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) SOLUTION: 6. (AA Sim.) Given: is a semicircle. 7. (Def. of

Prove: COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each

Proof: side. Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.) Since , triangle BCD is a right triangle. 7. (Def. of So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to COORDINATE GEOMETRY Find the Write an equation using the cosine ratio. measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio.

50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – If 5), and D(–1, 2) Use a calculator. SOLUTION: Use the Distance Formula to find the length of each side. 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2)

SOLUTION: Use the Distance Formula to find the length of each side.

Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio. In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

If Write an equation using the sine ratio.

Use a calculator.

If 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – Use a calculator. 2), and Z(7, –2)

SOLUTION: Use the Distance Formula to find the length of each Solve each equation. side. 52. x2 + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4. In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite 2 53. x – 6x = –9 Write an equation using the sine ratio. SOLUTION:

Use a calculator. Therefore, the solution is 3.

Solve each equation. 54. 3x2 + 15x = 0 2 52. x + 13x = –36 SOLUTION: SOLUTION:

Therefore, the solution is –5, 0.

Therefore, the solution is –9, –4. 2 55. 28 = x + 3x 2 53. x – 6x = –9 SOLUTION: SOLUTION:

Therefore, the solution is –7, 4. Therefore, the solution is 3. 56. x2 + 12x + 36 = 0 2 54. 3x + 15x = 0 SOLUTION: SOLUTION:

Therefore, the solution is –6. Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144. Find each measure. Assume that segments that appear to be tangent are tangent. 3. 1.

SOLUTION: SOLUTION:

2. 4.

SOLUTION: SOLUTION:

So, the measure of arc TS is 144.

3. 5.

SOLUTION: SOLUTION:

So, the measure of arc QTS is 248. 4. 6.

SOLUTION: SOLUTION:

So, the measure of arc LP is 150. 5. 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

SOLUTION: So, the measure of arc QTS is 248. Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting 6. tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that So, the measure of arc LP is 150. appear to be tangent are tangent. 8. 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting 9. tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 10. 8.

SOLUTION: SOLUTION:

∠JMK and ∠HMJ form a linear pair.

9. 11.

SOLUTION:

10. Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

∠JMK and ∠HMJ form a linear pair.

13. 11.

SOLUTION: SOLUTION:

So, the measure of arc PM is 144.

Therefore, the measure of arc RQ is 28. 14. 12.

SOLUTION: Arc BD and arc BCD are a minor and major arc that SOLUTION: share the same endpoints.

13. 15.

SOLUTION:

By Theorem 10.13, . So, the measure of arc PM is 144.

14.

16.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

By Theorem 11.13, . 15. Solve for .

We know that . Substitute.

SOLUTION: 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , By Theorem 10.13, . find each measure.

16.

a. b. SOLUTION:

a. By Theorem 11.13, .

SOLUTION: Substitute.

By Theorem 11.13, . Simplify. Solve for . b. By Theorem 11.14,

We know that . . Substitute. Substitute.

Simplify. 17. SPORTS The multi-sport field shown includes a

softball field and a soccer field. If , find each measure.

CCSS STRUCTURE Find each measure. 18.

a. b. SOLUTION:

SOLUTION: By Theorem 11.14, .

a. By Theorem 11.13, . Substitute.

Substitute.

Simplify.

b. By Theorem 11.14,

. Substitute. 19.

Simplify.

CCSS STRUCTURE Find each measure.

18. SOLUTION:

By Theorem 11.14, . Substitute.

Simplify. SOLUTION: By Theorem 11.14, . Substitute.

Simplify. 20. m(arc JNM)

19.

SOLUTION:

SOLUTION: So, the measure of arc JNM is 205. By Theorem 11.14, .

Substitute. 21.

Simplify.

SOLUTION:

20. m(arc JNM) By Theorem 11.14, . Substitute.

Simplify.

22.

SOLUTION:

SOLUTION: So, the measure of arc JNM is 205. By Theorem 11.14, . Substitute. 21.

Simplify.

23. SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify. By Theorem 11.14, . Substitute.

22. Solve for .

24. JEWELRY In the circular necklace shown, A and B SOLUTION: are tangent points. If x = 260, what is y? By Theorem 11.14, . Substitute.

Simplify.

23. SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

By Theorem 11.14, . Substitute.

25. SPACE A satellite orbits above Earth’s equator. Solve for . Find x, the measure of the planet’s arc, that is visible

to the satellite.

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

’ Therefore, the measure of the planet s arc that is visible to the satellite is 168. SOLUTION: ALGEBRA Find the value of x. By Theorem 11.14, . Substitute.

26. Simplify. SOLUTION: By Theorem 11.14, . Solve for . 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

27.

Therefore, the measure of the planet’s arc that is SOLUTION: visible to the satellite is 168. By Theorem 11.14, . ALGEBRA Find the value of x. Solve for .

26. SOLUTION:

By Theorem 11.14, . Solve for .

28. SOLUTION:

By Theorem 11.14, . Solve for .

27. SOLUTION:

By Theorem 11.14, . 29. PHOTOGRAPHY A photographer frames a Solve for . carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

28. SOLUTION:

By Theorem 11.14, . Solve for . SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

29. PHOTOGRAPHY A photographer frames a b. Let x be the measure of the camera’s viewing carousel in his camera shot as shown so that the lines angle. of sight form tangents to the carousel. By Theorem 11.14, . a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? Solve for . b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove:

SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . SOLUTION: Solve for . Statements (Reasons)

1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its b. Let x be the measure of the camera’s viewing intercepted arc.) angle. 3. (Exterior

By Theorem 11.14, . Theorem)

Solve for . 4. (Substitution)

5. (Subtraction Prop.)

11-6 Secants, Tangents, and Angle Measures 6. (Distributive Prop.)

CCSS ARGUMENTS For each case of 31. Case 2 Theorem 11.14, write a two-column proof. Given: tangent and secant 30. Case 1 Given: secants and Prove:

Prove:

SOLUTION: Statements (Reasons) SOLUTION: Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given) 1. and are secants to the circle. (Given) 2. , ( The 2. , (The meas. of an inscribed the measure of its measure of an inscribed the measure of its intercepted arc.) intercepted arc.) 3. (Exterior 3. (Exterior Theorem) Theorem) 4. (Substitution) 4. (Substitution)

5. (Subtraction Prop.) 5. (Subtraction Prop.)

6. (Distributive Prop.) 6. (Distributive Prop.) 32. Case 3 31. Case 2 Given: tangent and Given: tangent and secant Prove:

Prove:

SOLUTION: SOLUTION: Statements (Reasons) Statements (Reasons) 1. is a tangent to the circle and is a secant to 1. and are tangents to the circle. (Given) the circle. (Given) 2. (The meas. 2. , ( The of an secant-tangent the measure of its meas. of an inscribed the measure of its intercepted arc.) intercepted arc.) 3. (Exterior eSolutions3. Manual - Powered by Cognero (Exterior Theorem) Page 8 Theorem) 4. (Substitution) 4. (Substitution)

5. (Subtraction Prop.) 5. (Subtraction Prop.)

6. (Distributive Prop.) 6. (Distributive Prop.) 33. PROOF Write a paragraph proof of Theorem 11.13. 32. Case 3 Given: tangent and

Prove:

SOLUTION: a. Given: is a tangent of . Statements (Reasons) is a secant of . 1. and are tangents to the circle. (Given) ∠CAE is acute. Prove: m∠CAE = m(arc CA) 2. (The meas. b. Prove that if ∠CAB is obtuse, m∠CAB = m of an secant-tangent the measure of its (arc CDA) intercepted arc.) 3. (Exterior SOLUTION: Theorem) a. Proof: By Theorem 11.10, . So, ∠FAE

4. (Substitution) is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is

5. (Subtraction Prop.) acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC +

6. (Distributive Prop.) m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = 33. PROOF Write a paragraph proof of Theorem 11.13. m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution.

m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m

(arc FC) + m(arc CA). By Subtraction Prop.,

m∠CAE = m(arc CA).

b. Proof: Using the Angle and Arc Addition a. Given: is a tangent of . Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar is a secant of . CDA) = m(arc CF) + m(arc FDA). Since ∠CAE is acute. and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a Prove: m∠CAE = m(arc CA) measure of 180. By substitution, m∠CAB = m∠CAF b. Prove that if ∠CAB is obtuse, m∠CAB = m + 90 and m(arc CDA) = m(arc CF) + 180. Since (arc CDA) ∠CAF is inscribed, m∠CAF = m(arc CF) and by substitution, m∠CAB = m(arc CF) + 90. Using the SOLUTION: Division and Subtraction Properties on the Arc a. Proof: By Theorem 11.10, . So, ∠FAE Addition equation yields m(arc CDA) – m(arc is a right with measure of 90 and arc FCA is a ∠ CF) = 90. By substituting for 90, m∠CAB = m(arc semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle CF) + m(arc CDA) – m(arc CF). Then by and Arc Addition Postulates, m∠FAE = m∠FAC + subtraction, m∠CAB = m(arc CDA). m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution. m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m 34. WALLPAPER The design shown is an example of optical wallpaper. is a diameter of If m A (arc FC) + m(arc CA). By Subtraction Prop., ∠ = 26 and = 67, what is m CAE = m(arc CA). ∠ Refer to the image on page 766.

b. Proof: Using the Angle and Arc Addition SOLUTION: Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar First, find the measure of arc BD. CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a Since is a diameter, arc BDE is a semicircle and measure of 180. By substitution, m∠CAB = m∠CAF has a measure of 180. + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the MULTIPLE REPRESENTATIONS Division and Subtraction Properties on the Arc 35. In this problem, you will explore the relationship between Addition equation yields m(arc CDA) – m(arc Theorems 11.12 and 11.6. CF) = 90. By substituting for 90, m∠CAB = m(arc a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position CF) + m(arc CDA) – m(arc CF). Then by of point D moves closer to point C, but points A, B, and C remain fixed. subtraction, m∠CAB = m(arc CDA).

b. TABULAR Estimate the measure of for each successive circle, recording the measures of and in a table. Then calculate and record the value of x for each circle.

= 26 and = 67, what is Refer to the image on page 766. d. ANALYTICAL Write an algebraic proof to SOLUTION: show the relationship between Theorems 11.12 and First, find the measure of arc BD. 11.6 described in part c.

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

Theorems 11.12 and 11.6. Then draw chords and . a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position of point D moves closer to point C, but points A, B, and C remain fixed.

b. b. TABULAR Estimate the measure of for each successive circle, recording the measures of and in a table. Then calculate and record the value of x for each circle.

c. VERBAL Describe the relationship between and the value of x as approaches zero. c. As the measure of gets closer to 0, the What type of angle does ∠AEB become when measure of x approaches half of becomes an inscribed angle.

d. ANALYTICAL Write an algebraic proof to d. show the relationship between Theorems 11.12 and 11.6 described in part c. 36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: Find the difference of the two intercepted arcs and divide by 2. SOLUTION: 37. CHALLENGE The circles below are concentric. a. Redraw the circle with points A, B, and C in the What is x? same place but place point D closer to C each time. Then draw chords and .

b. SOLUTION:

c. As the measure of gets closer to 0, the measure of x approaches half of Use the arcs intercepted on the smaller circle to find

becomes an inscribed angle. m∠A.

divide by 2. 37. CHALLENGE The circles below are concentric. 38. REASONS Isosceles is inscribed in What is x? What can you conclude about and Explain.

SOLUTION: SOLUTION: Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC). Use the arcs intercepted on the smaller circle to find m∠A. 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain.

Use the arcs intercepted on the larger circle and b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain. m∠A to find the value of x.

38. REASONS Isosceles is inscribed in SOLUTION: What can you conclude about and a. ; for all values except when

Explain. at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to

the answer. SOLUTION: 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to Sample answer: m∠BAC = m∠BCA measure the angle that is formed. Find the measures because the triangle is isosceles. Since ∠BAC and of the minor and major arcs formed. Explain your ∠BCA are inscribed angles, by Theorem 11.6, m reasoning. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC). SOLUTION: Sample answer:

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ By Theorem 11.13, So, 50° = and KH. Explain.

Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°.

41. WRITING IN MATH A circle is inscribed within SOLUTION: If m∠P = 50 and m∠Q = 60, describe how a. ; for all values except when to find the measures of the three minor arcs formed by the points of tangency. at G, then . SOLUTION: b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to Sample answer: Use Theorem 11.14 to find each measure the angle that is formed. Find the measures minor arc. of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer:

The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70. By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130, Therefore, the measures of the three minor arcs are and y (major arc) = 360° – 130° or 230°. 130, 120, and 110.

41. WRITING IN MATH A circle is inscribed within 42. What is the value of x if and If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. SOLUTION:

A 23° B 31° C 64° D 128°

Sample answer: Use Theorem 11.14 to find each SOLUTION: minor arc. By Theorem 11.14, . Substitute.

Simplify.

The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70.

Therefore, the measures of the three minor arcs are 130, 120, and 110. So, the correct choice is C. 42. What is the value of x if and 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) A 23° J (1, 5) B 31° SOLUTION: C 64°

D 128° Here, C is the midpoint of . Use the midpoint formula, SOLUTION: , to find the coordinates of By Theorem 11.14, . C.

Substitute.

Simplify. So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

So, the correct choice is C. SOLUTION:

Here, C is the midpoint of . Use the midpoint formula, , to find the coordinates of Since , . C. 45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

So, the correct choice is J.

GRIDDED RESPONSE 44. If m∠AED = 95 and 2 A 64π units what is m∠BAC? B 32π units2 2 C 12π units D 8π units2 2 D 2π units

SOLUTION: SOLUTION: First, use the circumference to find the radius of the

Since . circle. We know that vertical angles are congruent. So, . Since , .

We know that the sum of the measures of all interior The shaded regions of the circle comprise half of the

angles of a triangle is 180. circle, so its area is half the area of the circle.

Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded The area of the shaded regions is 32π. regions? So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent.

2 A 64π units B 32π units2 2 C 12π units 46. D 8π units2 SOLUTION: 2 D 2π units By Theorem 11.10, . So, is a right triangle. SOLUTION: Use the Pythagorean Theorem. First, use the circumference to find the radius of the circle. Substitute.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

47. SOLUTION: If two segments from the same exterior point are The area of the shaded regions is 32π. tangent to a circle, then they are congruent. So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent.

48. 46. SOLUTION: SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem. By Theorem 11.10, . So, is a right triangle. Solve for x. Use the Pythagorean Theorem.

Substitute.

49. PROOF Write a two-column proof. Given: is a semicircle;

Prove:

47.

SOLUTION: If two segments from the same exterior point are SOLUTION: tangent to a circle, then they are congruent. Given: is a semicircle.

Prove:

48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Proof: Solve for x. Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .)

49. PROOF Write a two-column proof. 5. (Reflexive Prop.) 6. (AA Sim.) Given: is a semicircle; 7. (Def. of Prove: COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: SOLUTION: Given: is a semicircle. Use the Distance Formula to find the length of each side.

Prove:

Since , triangle BCD is a right Proof: triangle. Statements (Reasons) So, CD is the length of the hypotenuse and BC is the 1. MHT is a semicircle; (Given) length of the leg adjacent to

2. is a right angle. (If an inscribed angle Write an equation using the cosine ratio.

intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) If 5. (Reflexive Prop.) 6. (AA Sim.) Use a calculator.

7. (Def. of

COORDINATE GEOMETRY Find the 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, –

measure of each angle to the nearest tenth of a 2), and Z(7, –2) degree by using the Distance Formula and an SOLUTION: inverse trigonometric ratio. Use the Distance Formula to find the length of each 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – side. 5), and D(–1, 2)

SOLUTION: Use the Distance Formula to find the length of each side.

In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio.

Since , triangle BCD is a right triangle.

So, CD is the length of the hypotenuse and BC is the If length of the leg adjacent to Write an equation using the cosine ratio. Use a calculator.

Solve each equation. If 52. x2 + 13x = –36 Use a calculator. SOLUTION:

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) Therefore, the solution is –9, –4. SOLUTION: Use the Distance Formula to find the length of each 2 side. 53. x – 6x = –9 SOLUTION:

In right triangle XYZ, ZX is the length of the Therefore, the solution is 3. hypotenuse and YZ is the length of the leg opposite 54. 3x2 + 15x = 0 Write an equation using the sine ratio. SOLUTION:

If Therefore, the solution is –5, 0.

Use a calculator.

2 Solve each equation. 55. 28 = x + 3x 2 52. x + 13x = –36 SOLUTION: SOLUTION:

Therefore, the solution is –7, 4. Therefore, the solution is –9, –4. 56. x2 + 12x + 36 = 0 2 53. x – 6x = –9 SOLUTION: SOLUTION:

Therefore, the solution is –6.

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

SOLUTION: So, the measure of arc QTS is 248.

So, the measure of arc TS is 144. 6.

SOLUTION:

SOLUTION: So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION: SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165

5. or 195.

Therefore, the measure of the angle the ramp makes with the ground is 15.

SOLUTION: Find each measure. Assume that segments that appear to be tangent are tangent. 8.

So, the measure of arc QTS is 248.

6. SOLUTION:

SOLUTION: 9.

So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of SOLUTION: the angle the ramp makes with the ground?

10.

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc SOLUTION: with the same endpoints, so its measure is 360 – 165 or 195.

∠JMK and ∠HMJ form a linear pair.

Therefore, the measure of the angle the ramp makes with the ground is 15. 11. Find each measure. Assume that segments that appear to be tangent are tangent. 8.

SOLUTION:

Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

13.

10.

SOLUTION:

SOLUTION: So, the measure of arc PM is 144.

14.

∠JMK and ∠HMJ form a linear pair.

11. SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

15.

Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

By Theorem 10.13, .

SOLUTION:

16. 13.

SOLUTION: SOLUTION:

By Theorem 11.13, .

So, the measure of arc PM is 144. Solve for .

14. We know that . Substitute.

17. SPORTS The multi-sport field shown includes a

softball field and a soccer field. If , SOLUTION: find each measure. Arc BD and arc BCD are a minor and major arc that share the same endpoints.

15. a. b. SOLUTION:

a. By Theorem 11.13, . Substitute.

SOLUTION: Simplify. By Theorem 10.13, . b. By Theorem 11.14,

. Substitute.

16. Simplify.

CCSS STRUCTURE Find each measure. 18.

SOLUTION:

By Theorem 11.13, .

Solve for . SOLUTION:

We know that . By Theorem 11.14, . Substitute. Substitute.

17. SPORTS The multi-sport field shown includes a Simplify.

softball field and a soccer field. If , find each measure.

19.

a. b.

SOLUTION:

a. By Theorem 11.13, . SOLUTION: Substitute. By Theorem 11.14, .

Substitute.

Simplify.

Simplify. b. By Theorem 11.14,

. Substitute.

Simplify. 20. m(arc JNM)

CCSS STRUCTURE Find each measure. 18.

SOLUTION: SOLUTION:

By Theorem 11.14, .

Substitute. So, the measure of arc JNM is 205.

Simplify. 21.

19. SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION:

By Theorem 11.14, . Substitute. 22.

Simplify.

SOLUTION:

By Theorem 11.14, . Substitute. 20. m(arc JNM)

Simplify.

23.

SOLUTION:

By Theorem 11.14, . So, the measure of arc JNM is 205. Substitute.

21. Solve for .

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION:

22. By Theorem 11.14, . Substitute.

Simplify.

SOLUTION:

By Theorem 11.14, .

Substitute. 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite. Simplify.

23. SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

SOLUTION:

By Theorem 11.14, . Therefore, the measure of the planet’s arc that is Substitute. visible to the satellite is 168.

ALGEBRA Find the value of x.

Solve for .

26. 24. JEWELRY In the circular necklace shown, A and B SOLUTION: are tangent points. If x = 260, what is y? By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify. 27.

SOLUTION:

By Theorem 11.14, .

25. SPACE A satellite orbits above Earth’s equator. Solve for . Find x, the measure of the planet’s arc, that is visible to the satellite.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x. 28. SOLUTION:

By Theorem 11.14, .

Therefore, the measure of the planet’s arc that is Solve for . visible to the satellite is 168.

ALGEBRA Find the value of x.

26. SOLUTION:

By Theorem 11.14, . 29. PHOTOGRAPHY A photographer frames a . Solve for carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

27. SOLUTION: SOLUTION: a. By Theorem 11.14, . Let x be the measure of the carousel that appears in the shot. Solve for . By Theorem 11.14, . Solve for .

b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . 28. Solve for . SOLUTION: By Theorem 11.14, . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove:

29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the SOLUTION: arc measure of the carousel that appears in the shot? Statements (Reasons) b. If you want to capture an arc measure of in 1. and are secants to the circle. (Given) the photograph, what viewing angle should be used? 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) SOLUTION: 6. (Distributive Prop.) a. Let x be the measure of the carousel that appears in the shot. 31. Case 2 By Theorem 11.14, . Given: tangent and secant Solve for . Prove:

b. Let x be the measure of the camera’s viewing angle. SOLUTION:

By Theorem 11.14, . Statements (Reasons) 1. is a tangent to the circle and is a secant to Solve for . the circle. (Given)

2. , ( The

meas. of an inscribed the measure of its intercepted arc.) CCSS ARGUMENTS For each case of 3. (Exterior Theorem 11.14, write a two-column proof. Theorem)

30. Case 1 4. (Substitution) Given: secants and

Prove: 5. (Subtraction Prop.)

6. (Distributive Prop.)

32. Case 3 Given: tangent and

SOLUTION: Prove: Statements (Reasons) 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its

intercepted arc.) 3. (Exterior SOLUTION: Theorem) Statements (Reasons) 1. and are tangents to the circle. (Given) 4. (Substitution) 2. (The meas. 5. (Subtraction Prop.) of an secant-tangent the measure of its 6. (Distributive Prop.) intercepted arc.) 3. (Exterior 31. Case 2 Theorem)

Given: tangent and secant 4. (Substitution)

Prove: 5. (Subtraction Prop.)

6. (Distributive Prop.)

33. PROOF Write a paragraph proof of Theorem 11.13.

SOLUTION: Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given)

2. , ( The

meas. of an inscribed the measure of its

intercepted arc.) 3. (Exterior a. Given: is a tangent of . Theorem) is a secant of .

4. (Substitution) ∠CAE is acute.

Prove: m∠CAE = m(arc CA) 5. (Subtraction Prop.) b. Prove that if ∠CAB is obtuse, m∠CAB = m

11-66. Secants, Tangents, and Angle (Distributive Prop.) Measures (arc CDA)

32. Case 3 SOLUTION: Given: tangent and a. Proof: By Theorem 11.10, . So, ∠FAE is a right with measure of 90 and arc FCA is a Prove: ∠ semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) +

(arc CA) by Division Prop., and m∠FAC + m∠CAE SOLUTION: = m(arc FC) + m(arc CA) by substitution. Statements (Reasons) 1. and are tangents to the circle. (Given) m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m 2. (The meas. (arc FC) + m(arc CA). By Subtraction Prop., of an secant-tangent the measure of its m∠CAE = m(arc CA). intercepted arc.) 3. (Exterior b. Proof: Using the Angle and Arc Addition Theorem) Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since 4. (Substitution) and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a 5. (Subtraction Prop.) measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since 6. (Distributive Prop.) ∠CAF is inscribed, m∠CAF = m(arc CF) and by

33. PROOF Write a paragraph proof of Theorem 11.13. substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m∠CAB = m(arc

CF) + m(arc CDA) – m(arc CF). Then by subtraction, m∠CAB = m(arc CDA).

a. Given: is a tangent of . is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA) 34. WALLPAPER The design shown is an example of optical wallpaper. is a diameter of If m∠A b. Prove that if ∠CAB is obtuse, m∠CAB = m

(arc CDA) = 26 and = 67, what is Refer to the image on page 766. SOLUTION: eSolutionsSOLUTION: Manual - Powered by Cognero Page 9 First, find the measure of arc BD. a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle Since is a diameter, arc BDE is a semicircle and and Arc Addition Postulates, m∠FAE = m∠FAC + has a measure of 180. m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + 35. MULTIPLE REPRESENTATIONS In this (arc CA) by Division Prop., and m∠FAC + m∠CAE problem, you will explore the relationship between = m(arc FC) + m(arc CA) by substitution. Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then m∠FAC = m(arc FC) since ∠FAC is inscribed, so draw three successive figures in which the position substitution yields m(arc FC) + m∠CAE = m of point D moves closer to point C, but points A, B, and C remain fixed. (arc FC) + m(arc CA). By Subtraction Prop.,

m∠CAE = m(arc CA). b. TABULAR Estimate the measure of for each successive circle, recording the measures of b. Proof: Using the Angle and Arc Addition and in a table. Then calculate and record Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar the value of x for each circle. CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a c. VERBAL Describe the relationship between measure of 90 and arc FDA is a semicircle with a and the value of x as approaches zero. measure of 180. By substitution, m∠CAB = m∠CAF What type of angle does ∠AEB become when + 90 and m(arc CDA) = m(arc CF) + 180. Since

∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the d. ANALYTICAL Write an algebraic proof to Division and Subtraction Properties on the Arc show the relationship between Theorems 11.12 and

Addition equation yields m(arc CDA) – m(arc 11.6 described in part c.

CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by subtraction, m∠CAB = m(arc CDA).

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and .

34. WALLPAPER The design shown is an example of optical wallpaper. is a diameter of If m∠A = 26 and = 67, what is b. Refer to the image on page 766. SOLUTION: First, find the measure of arc BD.

Since is a diameter, arc BDE is a semicircle and c. As the measure of gets closer to 0, the has a measure of 180. measure of x approaches half of becomes an inscribed angle.

d. 35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between Theorems 11.12 and 11.6. 36. WRITING IN MATH Explain how to find the a. GEOMETRIC Copy the figure shown. Then measure of an angle formed by a secant and a draw three successive figures in which the position tangent that intersect outside a circle. of point D moves closer to point C, but points A, B, SOLUTION: and C remain fixed. Find the difference of the two intercepted arcs and divide by 2. b. TABULAR Estimate the measure of for each successive circle, recording the measures of 37. CHALLENGE The circles below are concentric. and in a table. Then calculate and record What is x? the value of x for each circle.

c. VERBAL Describe the relationship between and the value of x as approaches zero. What type of angle does ∠AEB become when

SOLUTION: d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c.

Use the arcs intercepted on the smaller circle to find m∠A.

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time.

Then draw chords and . Use the arcs intercepted on the larger circle and m∠A to find the value of x.

38. REASONS Isosceles is inscribed in What can you conclude about and Explain.

c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle.

d. SOLUTION: Sample answer: m∠BAC = m∠BCA 36. WRITING IN MATH Explain how to find the because the triangle is isosceles. Since ∠BAC and measure of an angle formed by a secant and a ∠BCA are inscribed angles, by Theorem 11.6, m tangent that intersect outside a circle. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. SOLUTION: So, m(arc AB) = m(arc BC). Find the difference of the two intercepted arcs and divide by 2. 39. CCSS ARGUMENTS In the figure, is a 37. CHALLENGE The circles below are concentric. diameter and is a tangent.

What is x? a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

SOLUTION: SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and

Use the arcs intercepted on the smaller circle to find x degrees. Hence solving leads to m∠A. the answer.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures

Use the arcs intercepted on the larger circle and of the minor and major arcs formed. Explain your reasoning. m∠A to find the value of x. SOLUTION: Sample answer:

38. REASONS Isosceles is inscribed in What can you conclude about and

Explain. By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°.

41. WRITING IN MATH A circle is inscribed within SOLUTION: If m∠P = 50 and m∠Q = 60, describe how Sample answer: m∠BAC = m∠BCA to find the measures of the three minor arcs formed by the points of tangency. because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m SOLUTION: (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC).

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m G. ∠ Sample answer: Use Theorem 11.14 to find each Explain. minor arc. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

SOLUTION: The sum of the angles of a triangle is 180, so m∠R = a. ; for all values except when 180 – (50 + 60) or 70. at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and

x degrees. Hence solving leads to Therefore, the measures of the three minor arcs are 130, 120, and 110. the answer. 42. What is the value of x if and OPEN ENDED 40. Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer: A 23° B 31° C 64° D 128° SOLUTION:

By Theorem 11.13, So, 50° = By Theorem 11.14, . Substitute. Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°. Simplify. 41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. SOLUTION:

So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on Sample answer: Use Theorem 11.14 to find each circle C, and is a diameter. What are the minor arc. coordinates of C? F (2, 10) G (10, –6) H (5, –3) J (1, 5) SOLUTION: Here, C is the midpoint of . The sum of the angles of a triangle is 180, so m∠R = Use the midpoint formula, 180 – (50 + 60) or 70. , to find the coordinates of C.

Therefore, the measures of the three minor arcs are 130, 120, and 110. So, the correct choice is J. 42. What is the value of x if and 44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

A 23° B 31° C 64° SOLUTION: D 128° Since . SOLUTION: We know that vertical angles are congruent. By Theorem 11.14, . So, . Since , Substitute. . We know that the sum of the measures of all interior angles of a triangle is 180. Simplify. Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on 2 circle C, and is a diameter. What are the A 64π units coordinates of C? B 32π units2 2 F (2, 10) C 12π units G (10, –6) D 8π units2 H (5, –3) 2 J (1, 5) D 2π units SOLUTION: SOLUTION: First, use the circumference to find the radius of the

Here, C is the midpoint of . circle. Use the midpoint formula, , to find the coordinates of C.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

The area of the shaded regions is 32π. So, the correct choice is B.

SOLUTION: Find x. Assume that segments that appear to be tangent are tangent.

Since . We know that vertical angles are congruent. So, .

Since , 46. . SOLUTION: We know that the sum of the measures of all interior By Theorem 11.10, . So, is a right angles of a triangle is 180. triangle. Use the Pythagorean Theorem. Substitute.

Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

2 A 64π units 47. B 32π units2 SOLUTION: 2 C 12π units If two segments from the same exterior point are 2 D 8π units tangent to a circle, then they are congruent. 2 D 2π units SOLUTION: First, use the circumference to find the radius of the circle.

48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. Solve for x.

49. PROOF Write a two-column proof.

The area of the shaded regions is 32π. Given: is a semicircle; So, the correct choice is B. Prove: Find x. Assume that segments that appear to be tangent are tangent.

46. SOLUTION: SOLUTION: Given: is a semicircle. By Theorem 11.10, . So, is a right

triangle. Use the Pythagorean Theorem. Prove:

Substitute.

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 47. 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) SOLUTION: 5. (Reflexive Prop.) If two segments from the same exterior point are 6. (AA Sim.) tangent to a circle, then they are congruent.

7. (Def. of

COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 48. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – SOLUTION: 5), and D(–1, 2) Use Theorem 11.10 and the Pythagorean Theorem. SOLUTION: Use the Distance Formula to find the length of each Solve for x. side.

49. PROOF Write a two-column proof. Given: is a semicircle; Since , triangle BCD is a right triangle. Prove: So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

If SOLUTION: Use a calculator. Given: is a semicircle.

Prove: 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Use the Distance Formula to find the length of each side.

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) In right triangle XYZ, ZX is the length of the 3. is a right angle (Def. of ⊥ lines) hypotenuse and YZ is the length of the leg opposite 4. (All rt. angles are .) 5. (Reflexive Prop.) Write an equation using the sine ratio. 6. (AA Sim.)

7. (Def. of If COORDINATE GEOMETRY Find the

measure of each angle to the nearest tenth of a Use a calculator. degree by using the Distance Formula and an inverse trigonometric ratio. Solve each equation.

50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 2 5), and D(–1, 2) 52. x + 13x = –36 SOLUTION: SOLUTION: Use the Distance Formula to find the length of each side.

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION: Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio. Therefore, the solution is 3.

54. 3x2 + 15x = 0 If SOLUTION: Use a calculator.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – Therefore, the solution is –5, 0.

2), and Z(7, –2) SOLUTION: 2 Use the Distance Formula to find the length of each 55. 28 = x + 3x side. SOLUTION:

Therefore, the solution is –7, 4. In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite 56. x2 + 12x + 36 = 0

SOLUTION: Write an equation using the sine ratio.

If Therefore, the solution is –6.

Use a calculator.

Solve each equation. 52. x2 + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

4. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc QTS is 248. So, the measure of arc TS is 144. 6. 3.

SOLUTION:

So, the measure of arc LP is 150.

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc 5. with the same endpoints, so its measure is 360 – 165 or 195.

Therefore, the measure of the angle the ramp makes

with the ground is 15. SOLUTION: Find each measure. Assume that segments that appear to be tangent are tangent. 8.

So, the measure of arc QTS is 248.

SOLUTION:

So, the measure of arc LP is 150.

10.

∠JMK and ∠HMJ form a linear pair.

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that 11. appear to be tangent are tangent. 8.

SOLUTION: SOLUTION:

Therefore, the measure of arc RQ is 28.

12.

SOLUTION: SOLUTION:

13.

10.

SOLUTION:

So, the measure of arc PM is 144.

14. ∠JMK and ∠HMJ form a linear pair.

11.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

15. Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

By Theorem 10.13, . SOLUTION:

13. 16.

SOLUTION:

By Theorem 11.13, . So, the measure of arc PM is 144. Solve for . 14. We know that . Substitute.

17. SPORTS The multi-sport field shown includes a SOLUTION: Arc BD and arc BCD are a minor and major arc that softball field and a soccer field. If ,

share the same endpoints. find each measure.

15.

a. b. SOLUTION:

a. By Theorem 11.13, .

Substitute. SOLUTION: By Theorem 10.13, . Simplify.

b. By Theorem 11.14,

. Substitute.

16. Simplify.

CCSS STRUCTURE Find each measure. 18. SOLUTION:

By Theorem 11.13, .

Solve for .

We know that . SOLUTION: Substitute. By Theorem 11.14, . Substitute.

17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , Simplify. find each measure.

19.

a. b. SOLUTION:

a. By Theorem 11.13, . Substitute. SOLUTION: By Theorem 11.14, .

Simplify. Substitute.

b. By Theorem 11.14, Simplify. . Substitute.

Simplify.

20. m(arc JNM)

CCSS STRUCTURE Find each measure. 18.

SOLUTION: SOLUTION:

By Theorem 11.14, . Substitute.

So, the measure of arc JNM is 205.

Simplify.

21.

19.

SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

By Theorem 11.14, . Substitute. 22.

Simplify.

SOLUTION:

By Theorem 11.14, . m(arc JNM) 20. Substitute.

Simplify.

23.

SOLUTION:

So, the measure of arc JNM is 205. By Theorem 11.14, .

Substitute.

21.

Solve for .

JEWELRY 24. In the circular necklace shown, A and B are tangent points. If x = 260, what is y? SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

22. SOLUTION: By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

By Theorem 11.14, . Substitute.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible Simplify. to the satellite.

23.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x. SOLUTION:

By Theorem 11.14, .

Substitute. Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x. Solve for .

24. JEWELRY In the circular necklace shown, A and B 26. are tangent points. If x = 260, what is y? SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify. 27. SOLUTION:

By Theorem 11.14, . 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible Solve for . to the satellite.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x. 28. SOLUTION:

Therefore, the measure of the planet’s arc that is By Theorem 11.14, . visible to the satellite is 168. Solve for .

ALGEBRA Find the value of x.

26. SOLUTION:

By Theorem 11.14, . Solve for . 29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

27. SOLUTION: By Theorem 11.14, . SOLUTION: a. Let x be the measure of the carousel that appears Solve for . in the shot.

By Theorem 11.14, . Solve for .

b. Let x be the measure of the camera’s viewing angle. 28. By Theorem 11.14, . SOLUTION: Solve for . By Theorem 11.14, . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove:

29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? SOLUTION: b. If you want to capture an arc measure of in Statements (Reasons)

the photograph, what viewing angle should be used? 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

SOLUTION: 5. (Subtraction Prop.) a. Let x be the measure of the carousel that appears in the shot. 6. (Distributive Prop.)

By Theorem 11.14, . 31. Case 2 Solve for . Given: tangent and secant

Prove:

b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . SOLUTION: Statements (Reasons) Solve for . 1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The

meas. of an inscribed the measure of its CCSS ARGUMENTS For each case of intercepted arc.) Theorem 11.14, write a two-column proof. 3. (Exterior 30. Case 1 Theorem) Given: secants and 4. (Substitution)

Prove: 5. (Subtraction Prop.)

6. (Distributive Prop.)

32. Case 3 SOLUTION: Given: tangent and Statements (Reasons) Prove: 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem) SOLUTION: Statements (Reasons) 4. (Substitution) 1. and are tangents to the circle. (Given)

5. (Subtraction Prop.) 2. (The meas.

6. (Distributive Prop.) of an secant-tangent the measure of its intercepted arc.) 31. Case 2 3. (Exterior Given: tangent and secant Theorem)

Prove: 4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

33. PROOF Write a paragraph proof of Theorem 11.13. SOLUTION:

Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The

meas. of an inscribed the measure of its

intercepted arc.) 3. (Exterior Theorem) a. Given : is a tangent of . 4. (Substitution) is a secant of . ∠CAE is acute. 5. (Subtraction Prop.) Prove: m∠CAE = m(arc CA)

6. (Distributive Prop.) b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA) 32. Case 3 Given: tangent and SOLUTION:

m(arc FC) + m(arc CA). So, 90 = m(arc FC) + SOLUTION: (arc CA) by Division Prop., and m∠FAC + m∠CAE Statements (Reasons) = m(arc FC) + m(arc CA) by substitution. 1. and are tangents to the circle. (Given) m∠FAC = m(arc FC) since ∠FAC is inscribed, so 2. (The meas. substitution yields m(arc FC) + m∠CAE = m of an secant-tangent the measure of its (arc FC) + m(arc CA). By Subtraction Prop., intercepted arc.) m∠CAE = m(arc CA). 3. (Exterior Theorem) b. Proof: Using the Angle and Arc Addition

4. (Substitution) Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since

5. (Subtraction Prop.) and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a

6. (Distributive Prop.) measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by 33. PROOF Write a paragraph proof of Theorem 11.13. substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by

subtraction, m CAB = m(arc CDA). ∠

a. Given: is a tangent of .

is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA) 34. WALLPAPER The design shown is an example of b. Prove that if ∠CAB is obtuse, m∠CAB = m optical wallpaper. is a diameter of If m A (arc CDA) ∠ = 26 and = 67, what is Refer to the image on page 766. SOLUTION: SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a First, find the measure of arc BD. semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + Since is a diameter, arc BDE is a semicircle and m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). has a measure of 180. By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE 35. MULTIPLE REPRESENTATIONS In this = m(arc FC) + m(arc CA) by substitution. problem, you will explore the relationship between m∠FAC = m(arc FC) since ∠FAC is inscribed, so Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then substitution yields m(arc FC) + m∠CAE = m draw three successive figures in which the position (arc FC) + m(arc CA). By Subtraction Prop., of point D moves closer to point C, but points A, B,

and C remain fixed. m∠CAE = m(arc CA).

b. TABULAR Estimate the measure of for b. Proof: Using the Angle and Arc Addition each successive circle, recording the measures of Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar and in a table. Then calculate and record CDA) = m(arc CF) + m(arc FDA). Since the value of x for each circle. and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a c. VERBAL Describe the relationship between measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since and the value of x as approaches zero. What type of angle does ∠AEB become when ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc d. ANALYTICAL Write an algebraic proof to Addition equation yields m(arc CDA) – m(arc show the relationship between Theorems 11.12 and 11.6 described in part c. CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by subtraction, m∠CAB = m(arc CDA).

SOLUTION: 11-6 Secants, Tangents, and Angle Measures a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and . 34. WALLPAPER The design shown is an example of optical wallpaper. is a diameter of If m∠A

= 26 and = 67, what is Refer to the image on page 766. b. SOLUTION: First, find the measure of arc BD.

Since is a diameter, arc BDE is a semicircle and has a measure of 180. c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle. 35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between d. Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then 36. WRITING IN MATH Explain how to find the draw three successive figures in which the position measure of an angle formed by a secant and a of point D moves closer to point C, but points A, B, tangent that intersect outside a circle. and C remain fixed. SOLUTION: Find the difference of the two intercepted arcs and b. TABULAR Estimate the measure of for divide by 2. each successive circle, recording the measures of and in a table. Then calculate and record 37. CHALLENGE The circles below are concentric. the value of x for each circle. What is x?

c. VERBAL Describe the relationship between and the value of x as approaches zero. What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to SOLUTION: show the relationship between Theorems 11.12 and 11.6 described in part c.

Use the arcs intercepted on the smaller circle to find m∠A. SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and .

Use the arcs intercepted on the larger circle and eSolutions Manual - Powered by Cognero Page 10 m∠A to find the value of x.

38. REASONS Isosceles is inscribed in What can you conclude about and Explain. c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle.

d. SOLUTION: 36. WRITING IN MATH Explain how to find the Sample answer: m∠BAC = m∠BCA measure of an angle formed by a secant and a because the triangle is isosceles. Since ∠BAC and tangent that intersect outside a circle. ∠BCA are inscribed angles, by Theorem 11.6, m SOLUTION: (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. Find the difference of the two intercepted arcs and So, m(arc AB) = m(arc BC). divide by 2.

37. CHALLENGE The circles below are concentric. 39. CCSS ARGUMENTS In the figure, is a What is x? diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

SOLUTION:

SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is Use the arcs intercepted on the smaller circle to find involved the intercepted arcs measure (180 – x) and m∠A. x degrees. Hence solving leads to the answer.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to Use the arcs intercepted on the larger circle and measure the angle that is formed. Find the measures m∠A to find the value of x. of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer:

38. REASONS Isosceles is inscribed in What can you conclude about and Explain.

By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°. SOLUTION: 41. WRITING IN MATH A circle is inscribed within Sample answer: m∠BAC = m∠BCA If m∠P = 50 and m∠Q = 60, describe how because the triangle is isosceles. Since ∠BAC and to find the measures of the three minor arcs formed ∠BCA are inscribed angles, by Theorem 11.6, m by the points of tangency. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. SOLUTION: So, m(arc AB) = m(arc BC).

SOLUTION: a. ; for all values except when The sum of the angles of a triangle is 180, so m∠R = at G, then . 180 – (50 + 60) or 70. b. Because a diameter is involved the intercepted arcs measure (180 – x) and

x degrees. Hence solving leads to Therefore, the measures of the three minor arcs are the answer. 130, 120, and 110. 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to 42. What is the value of x if and measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer:

A 23° B 31° C 64° D 128°

SOLUTION: By Theorem 11.13, So, 50° = By Theorem 11.14, . Therefore, x (minor arc) = 130, Substitute. and y (major arc) = 360° – 130° or 230°.

41. WRITING IN MATH A circle is inscribed within Simplify. If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. SOLUTION:

So, the correct choice is C.

Sample answer: Use Theorem 11.14 to find each 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on minor arc. circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) J (1, 5) SOLUTION: The sum of the angles of a triangle is 180, so m∠R = Here, C is the midpoint of . 180 – (50 + 60) or 70. Use the midpoint formula, , to find the coordinates of C.

Therefore, the measures of the three minor arcs are 130, 120, and 110.

42. What is the value of x if and So, the correct choice is J. 44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

A 23° B 31° C 64° D 128° SOLUTION:

SOLUTION: Since . By Theorem 11.14, . We know that vertical angles are congruent. So, . Substitute. Since , . We know that the sum of the measures of all interior

Simplify. angles of a triangle is 180.

Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions? So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the 2 coordinates of C? A 64π units F 2 (2, 10) B 32π units G (10, –6) 2 C 12π units H (5, –3) 2 J (1, 5) D 8π units 2 D 2π units SOLUTION: Here, C is the midpoint of . SOLUTION: Use the midpoint formula, First, use the circumference to find the radius of the circle. , to find the coordinates of C.

The shaded regions of the circle comprise half of the So, the correct choice is J. circle, so its area is half the area of the circle.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

π The area of the shaded regions is 32 . So, the correct choice is B. SOLUTION: Find x. Assume that segments that appear to be Since . tangent are tangent. We know that vertical angles are congruent. So, . Since , . 46. We know that the sum of the measures of all interior SOLUTION: angles of a triangle is 180. By Theorem 11.10, . So, is a right Substitute. triangle. Use the Pythagorean Theorem.

Substitute. Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

2 A 64π units 2 B 32π units 47. 2 C 12π units SOLUTION: D 8π units2 2 If two segments from the same exterior point are D 2π units tangent to a circle, then they are congruent. SOLUTION: First, use the circumference to find the radius of the circle.

48.

SOLUTION: The shaded regions of the circle comprise half of the Use Theorem 11.10 and the Pythagorean Theorem. circle, so its area is half the area of the circle. Solve for x.

The area of the shaded regions is 32π. 49. PROOF Write a two-column proof. So, the correct choice is B. Given: is a semicircle;

Find x. Assume that segments that appear to be Prove: tangent are tangent.

46. SOLUTION: SOLUTION: By Theorem 11.10, . So, is a right Given: is a semicircle. triangle.

Use the Pythagorean Theorem.

Prove: Substitute.

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle 47. intercepts a semicircle, the angle is a right angle.) SOLUTION: 3. is a right angle (Def. of ⊥ lines) If two segments from the same exterior point are 4. (All rt. angles are .) tangent to a circle, then they are congruent. 5. (Reflexive Prop.) 6. (AA Sim.)

7. (Def. of

COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an 48. inverse trigonometric ratio. SOLUTION: 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – Use Theorem 11.10 and the Pythagorean Theorem. 5), and D(–1, 2) SOLUTION: Solve for x. Use the Distance Formula to find the length of each side.

49. PROOF Write a two-column proof. Given: is a semicircle;

Prove: Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

SOLUTION: If Given: is a semicircle. Use a calculator.

Prove: 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Use the Distance Formula to find the length of each side.

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) In right triangle XYZ, ZX is the length of the 4. (All rt. angles are .) hypotenuse and YZ is the length of the leg opposite 5. (Reflexive Prop.) 6. (AA Sim.) Write an equation using the sine ratio.

7. (Def. of

COORDINATE GEOMETRY Find the If measure of each angle to the nearest tenth of a degree by using the Distance Formula and an Use a calculator. inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) Solve each equation. 52. x2 + 13x = –36 SOLUTION: Use the Distance Formula to find the length of each SOLUTION: side.

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 Since , triangle BCD is a right SOLUTION: triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

Therefore, the solution is 3.

If 54. 3x2 + 15x = 0 Use a calculator. SOLUTION:

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) Therefore, the solution is –5, 0. SOLUTION: Use the Distance Formula to find the length of each 2 side. 55. 28 = x + 3x SOLUTION:

In right triangle XYZ, ZX is the length of the Therefore, the solution is –7, 4. hypotenuse and YZ is the length of the leg opposite 56. x2 + 12x + 36 = 0 Write an equation using the sine ratio. SOLUTION:

Use a calculator. Therefore, the solution is –6.

Solve each equation. 52. x2 + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

4. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc QTS is 248. So, the measure of arc TS is 144. 6. 3.

SOLUTION:

So, the measure of arc LP is 150.

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc 5. with the same endpoints, so its measure is 360 – 165 or 195.

Therefore, the measure of the angle the ramp makes with the ground is 15. SOLUTION: Find each measure. Assume that segments that appear to be tangent are tangent. 8.

So, the measure of arc QTS is 248.

SOLUTION:

So, the measure of arc LP is 150.

10.

∠JMK and ∠HMJ form a linear pair.

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that 11. appear to be tangent are tangent. 8.

SOLUTION:

Therefore, the measure of arc RQ is 28.

12.

SOLUTION: SOLUTION:

13.

10.

SOLUTION:

So, the measure of arc PM is 144.

14. ∠JMK and ∠HMJ form a linear pair.

11.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

15. Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

By Theorem 10.13, . SOLUTION:

13. 16.

SOLUTION:

By Theorem 11.13, . So, the measure of arc PM is 144. Solve for . 14. We know that . Substitute.

17. SPORTS The multi-sport field shown includes a SOLUTION: softball field and a soccer field. If , Arc BD and arc BCD are a minor and major arc that find each measure. share the same endpoints.

15.

a. b. SOLUTION:

a. By Theorem 11.13, .

Substitute. SOLUTION: By Theorem 10.13, . Simplify.

b. By Theorem 11.14,

. Substitute.

16. Simplify.

CCSS STRUCTURE Find each measure.

18. SOLUTION:

By Theorem 11.13, .

Solve for .

We know that . SOLUTION: Substitute. By Theorem 11.14, . Substitute.

17. SPORTS The multi-sport field shown includes a

softball field and a soccer field. If , Simplify. find each measure.

19.

a. b. SOLUTION:

a. By Theorem 11.13, . SOLUTION: Substitute. By Theorem 11.14, .

Simplify. Substitute.

b. By Theorem 11.14, Simplify. . Substitute.

Simplify.

20. m(arc JNM)

CCSS STRUCTURE Find each measure. 18.

SOLUTION: SOLUTION:

By Theorem 11.14, . Substitute. So, the measure of arc JNM is 205.

Simplify. 21.

19.

SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

By Theorem 11.14, . Substitute. 22.

Simplify.

SOLUTION:

By Theorem 11.14, . 20. m(arc JNM) Substitute.

Simplify.

23.

SOLUTION:

So, the measure of arc JNM is 205. By Theorem 11.14, .

Substitute.

21.

Solve for .

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y? SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION: 22. By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

By Theorem 11.14, . Substitute.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible Simplify. to the satellite.

23.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x. SOLUTION:

By Theorem 11.14, .

Substitute. Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x. Solve for .

26. 24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y? SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify. 27. SOLUTION:

By Theorem 11.14, . 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible Solve for . to the satellite.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x. 28. SOLUTION:

Therefore, the measure of the planet’s arc that is By Theorem 11.14, . visible to the satellite is 168. Solve for .

ALGEBRA Find the value of x.

26. SOLUTION:

27. SOLUTION: SOLUTION: By Theorem 11.14, . a. Let x be the measure of the carousel that appears Solve for . in the shot.

By Theorem 11.14, . Solve for .

b. Let x be the measure of the camera’s viewing angle. 28. By Theorem 11.14, . SOLUTION: Solve for . By Theorem 11.14, . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove:

29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? SOLUTION: b. If you want to capture an arc measure of in Statements (Reasons) the photograph, what viewing angle should be used? 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) SOLUTION: a. Let x be the measure of the carousel that appears 6. (Distributive Prop.) in the shot.

By Theorem 11.14, . 31. Case 2 Solve for . Given: tangent and secant

Prove:

b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . SOLUTION: Statements (Reasons) Solve for . 1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The

meas. of an inscribed the measure of its

CCSS ARGUMENTS For each case of intercepted arc.) Theorem 11.14, write a two-column proof. 3. (Exterior 30. Case 1 Theorem)

Given: secants and 4. (Substitution)

Prove: 5. (Subtraction Prop.)

6. (Distributive Prop.)

32. Case 3

SOLUTION: Given: tangent and

Statements (Reasons) Prove: 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem) SOLUTION: Statements (Reasons) 4. (Substitution) 1. and are tangents to the circle. (Given)

5. (Subtraction Prop.) 2. (The meas.

6. (Distributive Prop.) of an secant-tangent the measure of its intercepted arc.) 31. Case 2 3. (Exterior Given: tangent and secant Theorem)

Prove: 4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

33. PROOF Write a paragraph proof of Theorem 11.13. SOLUTION:

Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The

meas. of an inscribed the measure of its

intercepted arc.) 3. (Exterior Theorem) a. Given: is a tangent of .

4. (Substitution) is a secant of . ∠CAE is acute. 5. (Subtraction Prop.) Prove: m∠CAE = m(arc CA)

6. (Distributive Prop.) b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA) 32. Case 3 Given: tangent and SOLUTION:

Prove: a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + SOLUTION: (arc CA) by Division Prop., and m∠FAC + m∠CAE Statements (Reasons) = m(arc FC) + m(arc CA) by substitution. 1. and are tangents to the circle. (Given) m∠FAC = m(arc FC) since ∠FAC is inscribed, so 2. (The meas. substitution yields m(arc FC) + m∠CAE = m of an secant-tangent the measure of its (arc FC) + m(arc CA). By Subtraction Prop., intercepted arc.) m∠CAE = m(arc CA). 3. (Exterior Theorem) b. Proof: Using the Angle and Arc Addition Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar 4. (Substitution) CDA) = m(arc CF) + m(arc FDA). Since

5. (Subtraction Prop.) and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF 6. (Distributive Prop.) + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by 33. PROOF Write a paragraph proof of Theorem 11.13. substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m CAB = m(arc ∠ CF) + m(arc CDA) – m(arc CF). Then by

subtraction, m∠CAB = m(arc CDA).

a. Given: is a tangent of . is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA) 34. WALLPAPER The design shown is an example of b. Prove that if ∠CAB is obtuse, m∠CAB = m optical wallpaper. is a diameter of If m∠A (arc CDA) = 26 and = 67, what is Refer to the image on page 766. SOLUTION: SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE First, find the measure of arc BD. is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + Since is a diameter, arc BDE is a semicircle and m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). has a measure of 180. By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE 35. MULTIPLE REPRESENTATIONS In this = m(arc FC) + m(arc CA) by substitution. problem, you will explore the relationship between m∠FAC = m(arc FC) since ∠FAC is inscribed, so Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then substitution yields m(arc FC) + m∠CAE = m draw three successive figures in which the position (arc FC) + m(arc CA). By Subtraction Prop., of point D moves closer to point C, but points A, B, and C remain fixed. m∠CAE = m(arc CA). b. TABULAR Estimate the measure of for b. Proof: Using the Angle and Arc Addition each successive circle, recording the measures of Postulates, m CAB = m CAF + m FAB and m(ar ∠ ∠ ∠ and in a table. Then calculate and record CDA) = m(arc CF) + m(arc FDA). Since the value of x for each circle. and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a c. VERBAL Describe the relationship between measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since and the value of x as approaches zero. What type of angle does ∠AEB become when ∠CAF is inscribed, m∠CAF = m(arc CF) and by substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc d. ANALYTICAL Write an algebraic proof to Addition equation yields m(arc CDA) – m(arc show the relationship between Theorems 11.12 and 11.6 described in part c. CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by subtraction, m∠CAB = m(arc CDA).

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and . 34. WALLPAPER The design shown is an example of optical wallpaper. is a diameter of If m∠A

= 26 and = 67, what is Refer to the image on page 766. b. SOLUTION: First, find the measure of arc BD.

Since is a diameter, arc BDE is a semicircle and has a measure of 180. c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle.

35. MULTIPLE REPRESENTATIONS In this

problem, you will explore the relationship between d. Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then 36. WRITING IN MATH Explain how to find the draw three successive figures in which the position measure of an angle formed by a secant and a of point D moves closer to point C, but points A, B, tangent that intersect outside a circle. and C remain fixed. SOLUTION: Find the difference of the two intercepted arcs and b. TABULAR Estimate the measure of for divide by 2. each successive circle, recording the measures of and in a table. Then calculate and record 37. CHALLENGE The circles below are concentric. the value of x for each circle. What is x?

c. VERBAL Describe the relationship between and the value of x as approaches zero. What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to SOLUTION: show the relationship between Theorems 11.12 and 11.6 described in part c.

Use the arcs intercepted on the smaller circle to find

m∠A. SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and .

Use the arcs intercepted on the larger circle and m∠A to find the value of x.

38. REASONS Isosceles is inscribed in What can you conclude about and

Explain. c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle.

d. SOLUTION: 36. WRITING IN MATH Explain how to find the Sample answer: m∠BAC = m∠BCA measure of an angle formed by a secant and a because the triangle is isosceles. Since ∠BAC and tangent that intersect outside a circle. ∠BCA are inscribed angles, by Theorem 11.6, m SOLUTION: (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. 11-6Find Secants, the difference Tangents, of theand two Angle intercepted Measures arcs and So, m(arc AB) = m(arc BC). divide by 2.

SOLUTION:

SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is Use the arcs intercepted on the smaller circle to find involved the intercepted arcs measure (180 – x) and

m∠A. x degrees. Hence solving leads to the answer.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to Use the arcs intercepted on the larger circle and measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your m∠A to find the value of x. reasoning. SOLUTION: Sample answer:

38. REASONS Isosceles is inscribed in What can you conclude about and

Explain. By Theorem 11.13, So, 50° =

39. CCSS ARGUMENTS In the figure, is a eSolutionsdiameterManual and- Powered is a tangent.by Cognero Page 11 a. Describe the range of possible values for m∠G. Explain. b. If m G = 34, find the measures of minor arcs HJ Sample answer: Use Theorem 11.14 to find each ∠ and KH. Explain. minor arc.

SOLUTION:

a. ; for all values except when The sum of the angles of a triangle is 180, so m∠R = at G, then . 180 – (50 + 60) or 70. b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to Therefore, the measures of the three minor arcs are the answer. 130, 120, and 110. 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to 42. What is the value of x if and measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer:

A 23° B 31° C 64° D 128°

SOLUTION: By Theorem 11.13, So, 50° = By Theorem 11.14, . Therefore, x (minor arc) = 130, Substitute. and y (major arc) = 360° – 130° or 230°.

41. WRITING IN MATH A circle is inscribed within Simplify. If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. SOLUTION:

So, the correct choice is C.

SOLUTION: The sum of the angles of a triangle is 180, so m∠R = Here, C is the midpoint of . 180 – (50 + 60) or 70. Use the midpoint formula, , to find the coordinates of C.

Therefore, the measures of the three minor arcs are 130, 120, and 110.

42. What is the value of x if and So, the correct choice is J. 44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

A 23° B 31°

C 64° D 128° SOLUTION:

SOLUTION: Since .

By Theorem 11.14, . We know that vertical angles are congruent. So, .

Substitute. Since ,

. We know that the sum of the measures of all interior Simplify. angles of a triangle is 180.

Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions? So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the 2 coordinates of C? A 64π units F (2, 10) B 32π units2 G (10, –6) 2 C 12π units H (5, –3) D 2 J (1, 5) 8π units 2 D 2π units SOLUTION: SOLUTION: Here, C is the midpoint of . Use the midpoint formula, First, use the circumference to find the radius of the circle. , to find the coordinates of C.

The shaded regions of the circle comprise half of the So, the correct choice is J. circle, so its area is half the area of the circle.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

The area of the shaded regions is 32π. So, the correct choice is B. SOLUTION: Find x. Assume that segments that appear to be Since . tangent are tangent. We know that vertical angles are congruent. So, . Since , . 46. We know that the sum of the measures of all interior SOLUTION: angles of a triangle is 180.

By Theorem 11.10, . So, is a right triangle. Substitute. Use the Pythagorean Theorem.

Substitute. Since , . 45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

2 A 64π units 2 B 32π units 47. C 2 12π units SOLUTION: 2 D 8π units If two segments from the same exterior point are 2 D 2π units tangent to a circle, then they are congruent.

SOLUTION: First, use the circumference to find the radius of the circle.

48. SOLUTION:

The shaded regions of the circle comprise half of the Use Theorem 11.10 and the Pythagorean Theorem. circle, so its area is half the area of the circle. Solve for x.

The area of the shaded regions is 32π. 49. PROOF Write a two-column proof. So, the correct choice is B. Given: is a semicircle;

Find x. Assume that segments that appear to be Prove: tangent are tangent.

46. SOLUTION: SOLUTION: By Theorem 11.10, . So, is a right Given: is a semicircle. triangle. Use the Pythagorean Theorem.

Prove: Substitute.

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle 47. intercepts a semicircle, the angle is a right angle.)

SOLUTION: 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) If two segments from the same exterior point are tangent to a circle, then they are congruent. 5. (Reflexive Prop.) 6. (AA Sim.)

7. (Def. of

49. PROOF Write a two-column proof. Given: is a semicircle; Since , triangle BCD is a right Prove: triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

SOLUTION: If Given: is a semicircle. Use a calculator.

Prove: 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Use the Distance Formula to find the length of each side.

Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) In right triangle XYZ, ZX is the length of the 4. (All rt. angles are .) hypotenuse and YZ is the length of the leg opposite 5. (Reflexive Prop.) 6. (AA Sim.) Write an equation using the sine ratio.

7. (Def. of

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 Since , triangle BCD is a right SOLUTION: triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

Therefore, the solution is 3.

If 54. 3x2 + 15x = 0 Use a calculator. SOLUTION:

If Therefore, the solution is –6. Use a calculator.

Solve each equation. 52. x2 + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144. Find each measure. Assume that segments that appear to be tangent are tangent. 3. 1.

SOLUTION: SOLUTION:

2. 4.

SOLUTION: SOLUTION:

So, the measure of arc TS is 144.

3. 5.

SOLUTION: SOLUTION:

So, the measure of arc QTS is 248. 4. 6.

SOLUTION: SOLUTION:

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 10. 8.

SOLUTION: SOLUTION:

∠JMK and ∠HMJ form a linear pair.

9. 11.

SOLUTION:

10. Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

∠JMK and ∠HMJ form a linear pair.

13. 11.

SOLUTION: SOLUTION:

So, the measure of arc PM is 144.

Therefore, the measure of arc RQ is 28. 14. 12.

SOLUTION: Arc BD and arc BCD are a minor and major arc that SOLUTION: share the same endpoints.

13. 15.

SOLUTION:

By Theorem 10.13, . So, the measure of arc PM is 144.

14.

16.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

By Theorem 11.13, . 15. Solve for .

We know that . Substitute.

SOLUTION: 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , By Theorem 10.13, . find each measure.

16.

a. b. SOLUTION:

a. By Theorem 11.13, .

SOLUTION: Substitute.

By Theorem 11.13, . Simplify. Solve for . b. By Theorem 11.14,

We know that . . Substitute. Substitute.

Simplify. SPORTS 17. The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

CCSS STRUCTURE Find each measure. 18.

a. b. SOLUTION:

SOLUTION: By Theorem 11.14, .

a. By Theorem 11.13, . Substitute.

Substitute.

Simplify.

b. By Theorem 11.14,

. Substitute. 19.

Simplify.

CCSS STRUCTURE Find each measure.

18. SOLUTION:

By Theorem 11.14, . Substitute.

Simplify. SOLUTION: By Theorem 11.14, . Substitute.

Simplify. 20. m(arc JNM)

19.

SOLUTION:

SOLUTION: So, the measure of arc JNM is 205.

By Theorem 11.14, .

Substitute. 21.

Simplify.

SOLUTION:

20. m(arc JNM) By Theorem 11.14, . Substitute.

Simplify.

22.

SOLUTION:

SOLUTION: So, the measure of arc JNM is 205. By Theorem 11.14, . Substitute. 21.

Simplify.

23. SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify. By Theorem 11.14, . Substitute.

22. Solve for .

24. JEWELRY In the circular necklace shown, A and B SOLUTION: are tangent points. If x = 260, what is y? By Theorem 11.14, . Substitute.

Simplify.

23. SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

By Theorem 11.14, . Substitute.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible Solve for . to the satellite.

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is

visible to the satellite is 168. SOLUTION: ALGEBRA Find the value of x. By Theorem 11.14, . Substitute.

26. Simplify. SOLUTION: By Theorem 11.14, . Solve for . 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

27. SOLUTION: Therefore, the measure of the planet’s arc that is visible to the satellite is 168. By Theorem 11.14, . ALGEBRA Find the value of x. Solve for .

26. SOLUTION:

By Theorem 11.14, . Solve for .

28. SOLUTION:

By Theorem 11.14, . Solve for .

27. SOLUTION:

28. SOLUTION:

By Theorem 11.14, . Solve for . SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove:

SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . SOLUTION: Solve for . Statements (Reasons)

1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its b. Let x be the measure of the camera’s viewing intercepted arc.)

angle. 3. (Exterior

By Theorem 11.14, . Theorem)

Solve for . 4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

CCSS ARGUMENTS For each case of 31. Case 2 Theorem 11.14, write a two-column proof. Given: tangent and secant 30. Case 1 Given: secants and Prove:

Prove:

intercepted arc.) 3. (Exterior 3. (Exterior Theorem) Theorem) 4. (Substitution) 4. (Substitution)

5. (Subtraction Prop.) 5. (Subtraction Prop.)

6. (Distributive Prop.) 6. (Distributive Prop.) 32. Case 3 31. Case 2 Given: tangent and

Given: tangent and secant Prove:

Prove:

SOLUTION: SOLUTION: Statements (Reasons) Statements (Reasons) 1. is a tangent to the circle and is a secant to 1. and are tangents to the circle. (Given) the circle. (Given) 2. (The meas. 2. , ( The of an secant-tangent the measure of its meas. of an inscribed the measure of its intercepted arc.) intercepted arc.) 3. (Exterior 3. (Exterior Theorem) Theorem) 4. (Substitution) 4. (Substitution)

5. (Subtraction Prop.) 5. (Subtraction Prop.)

6. (Distributive Prop.) 6. (Distributive Prop.) 33. PROOF Write a paragraph proof of Theorem 11.13. 32. Case 3 Given: tangent and

Prove:

Theorem) a. Proof: By Theorem 11.10, . So, ∠FAE

4. (Substitution) is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is

5. (Subtraction Prop.) acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC +

m FAC = m(arc FC) since FAC is inscribed, so ∠ ∠ substitution yields m(arc FC) + m∠CAE = m (arc FC) + m(arc CA). By Subtraction Prop.,

m∠CAE = m(arc CA).

b. Proof: Using the Angle and Arc Addition a. Given: is a tangent of . Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar is a secant of . CDA) = m(arc CF) + m(arc FDA). Since ∠CAE is acute. and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a Prove: m CAE = m(arc CA) ∠ measure of 180. By substitution, m∠CAB = m∠CAF b. Prove that if ∠CAB is obtuse, m∠CAB = m + 90 and m(arc CDA) = m(arc CF) + 180. Since (arc CDA) ∠CAF is inscribed, m∠CAF = m(arc CF) and by substitution, m∠CAB = m(arc CF) + 90. Using the SOLUTION: Division and Subtraction Properties on the Arc a. Proof: By Theorem 11.10, . So, ∠FAE Addition equation yields m(arc CDA) – m(arc is a right ∠ with measure of 90 and arc FCA is a CF) = 90. By substituting for 90, m∠CAB = m(arc semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle CF) + m(arc CDA) – m(arc CF). Then by and Arc Addition Postulates, m∠FAE = m∠FAC + subtraction, m∠CAB = m(arc CDA). m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution. m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m 34. WALLPAPER The design shown is an example of optical wallpaper. is a diameter of If m∠A (arc FC) + m(arc CA). By Subtraction Prop.,

= 26 and = 67, what is m∠CAE = m(arc CA). Refer to the image on page 766.

substitution, m∠CAB = m(arc CF) + 90. Using the 35. MULTIPLE REPRESENTATIONS In this Division and Subtraction Properties on the Arc problem, you will explore the relationship between Addition equation yields m(arc CDA) – m(arc Theorems 11.12 and 11.6. CF) = 90. By substituting for 90, m∠CAB = m(arc a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position CF) + m(arc CDA) – m(arc CF). Then by of point D moves closer to point C, but points A, B, and C remain fixed. subtraction, m∠CAB = m(arc CDA).

b. TABULAR Estimate the measure of for each successive circle, recording the measures of and in a table. Then calculate and record the value of x for each circle.

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

SOLUTION: 35. MULTIPLE REPRESENTATIONS In this a. Redraw the circle with points A, B, and C in the problem, you will explore the relationship between same place but place point D closer to C each time. Theorems 11.12 and 11.6. Then draw chords and . a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position of point D moves closer to point C, but points A, B,

and C remain fixed. b. b. TABULAR Estimate the measure of for each successive circle, recording the measures of and in a table. Then calculate and record the value of x for each circle.

d. ANALYTICAL Write an algebraic proof to d. show the relationship between Theorems 11.12 and 11.6 described in part c. WRITING IN MATH 36. Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: Find the difference of the two intercepted arcs and divide by 2. SOLUTION: 37. CHALLENGE The circles below are concentric. a. Redraw the circle with points A, B, and C in the What is x? same place but place point D closer to C each time. Then draw chords and .

b. SOLUTION:

c. As the measure of gets closer to 0, the measure of x approaches half of Use the arcs intercepted on the smaller circle to find m A. becomes an inscribed angle. ∠

36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a Use the arcs intercepted on the larger circle and tangent that intersect outside a circle. m∠A to find the value of x. SOLUTION: Find the difference of the two intercepted arcs and divide by 2. CHALLENGE 37. The circles below are concentric. 38. REASONS Isosceles is inscribed in What is x? What can you conclude about and Explain.

Use the arcs intercepted on the larger circle and b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain. m∠A to find the value of x.

38. REASONS Isosceles is inscribed in SOLUTION: What can you conclude about and a. ; for all values except when

Explain. at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer.

SOLUTION: 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to Sample answer: m BAC = m BCA ∠ ∠ measure the angle that is formed. Find the measures because the triangle is isosceles. Since ∠BAC and of the minor and major arcs formed. Explain your ∠BCA are inscribed angles, by Theorem 11.6, m reasoning. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC). SOLUTION: Sample answer:

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m G = 34, find the measures of minor arcs HJ ∠ By Theorem 11.13, So, 50° = and KH. Explain. Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°.

41. WRITING IN MATH A circle is inscribed within SOLUTION: If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed a. ; for all values except when by the points of tangency. at G, then . SOLUTION: b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer.

40. OPEN ENDED Draw a circle and two tangents

that intersect outside the circle. Use a protractor to Sample answer: Use Theorem 11.14 to find each measure the angle that is formed. Find the measures minor arc. of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer:

The sum of the angles of a triangle is 180, so m∠R =

180 – (50 + 60) or 70. By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130, 11-6 Secants, Tangents, and Angle Measures Therefore, the measures of the three minor arcs are and y (major arc) = 360° – 130° or 230°. 130, 120, and 110.

A 23° B 31° C 64° D 128°

Sample answer: Use Theorem 11.14 to find each SOLUTION: minor arc. By Theorem 11.14, . Substitute.

Simplify.

The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70.

Substitute.

Simplify. eSolutions Manual - Powered by Cognero So, the correct choice is J. Page 12 44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

So, the correct choice is C. SOLUTION:

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on Since . circle C, and is a diameter. What are the We know that vertical angles are congruent. coordinates of C? So, . F (2, 10) Since , G (10, –6) . H (5, –3) We know that the sum of the measures of all interior J (1, 5) angles of a triangle is 180. SOLUTION: Substitute. Here, C is the midpoint of . Use the midpoint formula, , to find the coordinates of Since , . C. 45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and 2 A 64π units what is m∠BAC? B 32π units2 2 C 12π units D 8π units2 2 D 2π units

SOLUTION: SOLUTION: First, use the circumference to find the radius of the circle. Since . We know that vertical angles are congruent. So, . Since , .

We know that the sum of the measures of all interior The shaded regions of the circle comprise half of the angles of a triangle is 180. circle, so its area is half the area of the circle.

Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below

is 16π units, what is the total area of the shaded The area of the shaded regions is 32π. regions? So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent.

2 A 64π units B 32π units2 2 46. C 12π units D 8π units2 SOLUTION: 2 D 2π units By Theorem 11.10, . So, is a right triangle. SOLUTION: Use the Pythagorean Theorem. First, use the circumference to find the radius of the circle. Substitute.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

47. SOLUTION: If two segments from the same exterior point are The area of the shaded regions is 32π. tangent to a circle, then they are congruent. So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent.

48. 46. SOLUTION: SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

By Theorem 11.10, . So, is a right triangle. Solve for x. Use the Pythagorean Theorem.

Substitute.

49. PROOF Write a two-column proof. Given: is a semicircle;

Prove:

47.

SOLUTION: If two segments from the same exterior point are SOLUTION: tangent to a circle, then they are congruent. Given: is a semicircle.

Prove:

48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem. Proof: Solve for x. Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.)

3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 49. PROOF Write a two-column proof. 6. (AA Sim.) Given: is a semicircle; 7. (Def. of Prove: COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: SOLUTION: Given: is a semicircle. Use the Distance Formula to find the length of each side.

Prove:

Since , triangle BCD is a right Proof: triangle. Statements (Reasons) So, CD is the length of the hypotenuse and BC is the 1. MHT is a semicircle; (Given) length of the leg adjacent to Write an equation using the cosine ratio. 2. is a right angle. (If an inscribed angle

intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines)

4. (All rt. angles are .) If 5. (Reflexive Prop.) 6. (AA Sim.) Use a calculator.

7. (Def. of

51. X in right triangle XYZ with vertices X(2, 2), Y(2, – COORDINATE GEOMETRY Find the ∠ 2), and Z(7, –2) measure of each angle to the nearest tenth of a degree by using the Distance Formula and an SOLUTION: inverse trigonometric ratio. Use the Distance Formula to find the length of each 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – side. 5), and D(–1, 2)

SOLUTION: Use the Distance Formula to find the length of each side.

In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio.

Since , triangle BCD is a right triangle.

So, CD is the length of the hypotenuse and BC is the If length of the leg adjacent to Write an equation using the cosine ratio. Use a calculator.

Solve each equation.

If 52. x2 + 13x = –36

Use a calculator. SOLUTION:

In right triangle XYZ, ZX is the length of the Therefore, the solution is 3. hypotenuse and YZ is the length of the leg opposite 2 54. 3x + 15x = 0

Write an equation using the sine ratio. SOLUTION:

If Therefore, the solution is –5, 0. Use a calculator.

2 Solve each equation. 55. 28 = x + 3x 52. x2 + 13x = –36 SOLUTION: SOLUTION:

Therefore, the solution is –7, 4. Therefore, the solution is –9, –4. 56. x2 + 12x + 36 = 0 2 53. x – 6x = –9 SOLUTION: SOLUTION:

Therefore, the solution is –6.

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION: SOLUTION:

2. 5.

SOLUTION: SOLUTION: So, the measure of arc TS is 144.

So, the measure of arc QTS is 248.

SOLUTION:

4. So, the measure of arc LP is 150. 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15. So, the measure of arc QTS is 248. Find each measure. Assume that segments that appear to be tangent are tangent. 6. 8.

SOLUTION: SOLUTION:

So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus 9. motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting

tangents to the circle formed by the barrel. One arc 10. has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that ∠JMK and ∠HMJ form a linear pair. appear to be tangent are tangent. 8.

11.

SOLUTION:

9. Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

10.

13.

SOLUTION:

∠JMK and ∠HMJ form a linear pair.

So, the measure of arc PM is 144. 14. 11.

SOLUTION: SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

Therefore, the measure of arc RQ is 28.

12. 15.

SOLUTION:

By Theorem 10.13, .

13.

16. SOLUTION:

So, the measure of arc PM is 144.

14. SOLUTION:

By Theorem 11.13, .

Solve for .

We know that . SOLUTION: Substitute. Arc BD and arc BCD are a minor and major arc that share the same endpoints.

17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

15.

a. SOLUTION: b. SOLUTION: By Theorem 10.13, .

a. By Theorem 11.13, . Substitute.

Simplify.

16. b. By Theorem 11.14,

. Substitute.

Simplify.

SOLUTION:

By Theorem 11.13, . CCSS STRUCTURE Find each measure. Solve for . 18.

We know that . Substitute.

17. SPORTS The multi-sport field shown includes a SOLUTION: softball field and a soccer field. If , By Theorem 11.14, . find each measure. Substitute.

Simplify.

a. b. 19. SOLUTION:

a. By Theorem 11.13, . Substitute.

Simplify.

SOLUTION: b. By Theorem 11.14,

. By Theorem 11.14, . Substitute. Substitute.

Simplify. Simplify.

CCSS STRUCTURE Find each measure. 18. 20. m(arc JNM)

SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

So, the measure of arc JNM is 205.

21. 19.

SOLUTION: SOLUTION: By Theorem 11.14, . By Theorem 11.14, . Substitute.

Substitute.

Simplify.

22.

20. m(arc JNM)

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION:

23.

So, the measure of arc JNM is 205.

21. SOLUTION:

By Theorem 11.14, . Substitute.

Solve for . SOLUTION:

By Theorem 11.14, . Substitute. 24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

Simplify.

22.

SOLUTION: SOLUTION: By Theorem 11.14, . By Theorem 11.14, . Substitute.

Substitute.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible 23. to the satellite.

SOLUTION: SOLUTION: By Theorem 11.14, . The measure of the visible arc is x and the measure Substitute. of the arc that is not visible is 360 – x. Use Theorem

11.14 to find the value of x.

Solve for .

Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

24. JEWELRY In the circular necklace shown, A and B ALGEBRA Find the value of x. are tangent points. If x = 260, what is y?

26. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible 27. to the satellite. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is visible to the satellite is 168. 28. ALGEBRA Find the value of x. SOLUTION:

By Theorem 11.14, . Solve for . 26. SOLUTION:

By Theorem 11.14, . Solve for .

27. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

28. SOLUTION:

By Theorem 11.14, . b. Let x be the measure of the camera’s viewing angle. Solve for . By Theorem 11.14, . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines Given: secants and of sight form tangents to the carousel. Prove: a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used?

SOLUTION: Statements (Reasons) 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) SOLUTION: 3. (Exterior a. Let x be the measure of the carousel that appears Theorem) in the shot. 4. (Substitution)

By Theorem 11.14, . 5. (Subtraction Prop.) Solve for .

6. (Distributive Prop.)

31. Case 2 Given: tangent and secant b. Let x be the measure of the camera’s viewing angle. Prove:

By Theorem 11.14, . Solve for .

SOLUTION: Statements (Reasons) 1. is a tangent to the circle and is a secant to CCSS ARGUMENTS For each case of the circle. (Given) Theorem 11.14, write a two-column proof. 2. , ( The 30. Case 1

Given: secants and meas. of an inscribed the measure of its

Prove: intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) SOLUTION: Statements (Reasons) 6. (Distributive Prop.) 1. and are secants to the circle. (Given) 2. , (The 32. Case 3 Given: tangent and measure of an inscribed the measure of its Prove: intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) SOLUTION: 6. (Distributive Prop.) Statements (Reasons) 1. and are tangents to the circle. (Given) 31. Case 2 2. (The meas. Given: tangent and secant

Prove: of an secant-tangent the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

SOLUTION: 5. (Subtraction Prop.) Statements (Reasons)

1. is a tangent to the circle and is a secant to 6. (Distributive Prop.) the circle. (Given) 2. , ( The 33. PROOF Write a paragraph proof of Theorem 11.13.

meas. of an inscribed the measure of its

intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.) a. Given: is a tangent of . is a secant of . 32. Case 3 ∠CAE is acute. Given: tangent and Prove: m∠CAE = m(arc CA)

Prove: b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA)

SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is SOLUTION: acute, C is in the interior of ∠FAE, so by the Angle Statements (Reasons) and Arc Addition Postulates, m∠FAE = m∠FAC + 1. and are tangents to the circle. (Given) m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = 2. (The meas. m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE of an secant-tangent the measure of its = m(arc FC) + m(arc CA) by substitution. intercepted arc.) 3. (Exterior m∠FAC = m(arc FC) since ∠FAC is inscribed, so Theorem) substitution yields m(arc FC) + m∠CAE = m

4. (Substitution) (arc FC) + m(arc CA). By Subtraction Prop., m∠CAE = m(arc CA). 5. (Subtraction Prop.)

b. Proof: Using the Angle and Arc Addition 6. (Distributive Prop.) Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since PROOF 33. Write a paragraph proof of Theorem 11.13. and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since

∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m CAB = m(arc ∠ a. Given: is a tangent of . CF) + m(arc CDA) – m(arc CF). Then by is a secant of . subtraction, m∠CAB = m(arc CDA). ∠CAE is acute. Prove: m∠CAE = m(arc CA)

b. Prove that if CAB is obtuse, m CAB = m ∠ ∠ (arc CDA)

SOLUTION: 34. WALLPAPER The design shown is an example of a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a optical wallpaper. is a diameter of If m∠A semicircle with measure of 180. Since ∠CAE is = 26 and = 67, what is acute, C is in the interior of ∠FAE, so by the Angle Refer to the image on page 766. and Arc Addition Postulates, m∠FAE = m∠FAC + SOLUTION: m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). First, find the measure of arc BD. By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE Since is a diameter, arc BDE is a semicircle and = m(arc FC) + m(arc CA) by substitution. has a measure of 180. m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m (arc FC) + m(arc CA). By Subtraction Prop., 35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between m∠CAE = m(arc CA). Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then b. Proof: Using the Angle and Arc Addition draw three successive figures in which the position Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar of point D moves closer to point C, but points A, B, CDA) = m(arc CF) + m(arc FDA). Since and C remain fixed. and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a b. TABULAR Estimate the measure of for measure of 180. By substitution, m∠CAB = m∠CAF each successive circle, recording the measures of + 90 and m(arc CDA) = m(arc CF) + 180. Since and in a table. Then calculate and record ∠CAF is inscribed, m∠CAF = m(arc CF) and by the value of x for each circle. substitution, m∠CAB = m(arc CF) + 90. Using the c. VERBAL Division and Subtraction Properties on the Arc Describe the relationship between Addition equation yields m(arc CDA) – m(arc and the value of x as approaches zero. What type of angle does ∠AEB become when CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by d. ANALYTICAL subtraction, m∠CAB = m(arc CDA). Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c.

b. Since is a diameter, arc BDE is a semicircle and has a measure of 180.

35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between c. As the measure of gets closer to 0, the Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then measure of x approaches half of draw three successive figures in which the position becomes an inscribed angle. of point D moves closer to point C, but points A, B, and C remain fixed. d.

b. TABULAR Estimate the measure of for 36. WRITING IN MATH Explain how to find the each successive circle, recording the measures of measure of an angle formed by a secant and a and in a table. Then calculate and record tangent that intersect outside a circle. the value of x for each circle. SOLUTION:

Find the difference of the two intercepted arcs and c. VERBAL Describe the relationship between divide by 2. and the value of x as approaches zero. What type of angle does ∠AEB become when 37. CHALLENGE The circles below are concentric. What is x?

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c.

SOLUTION:

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and . Use the arcs intercepted on the smaller circle to find m∠A.

Use the arcs intercepted on the larger circle and m∠A to find the value of x.

c. As the measure of gets closer to 0, the measure of x approaches half of 38. REASONS Isosceles is inscribed in becomes an inscribed angle. What can you conclude about and

Explain. d.

36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle.

SOLUTION: SOLUTION: Find the difference of the two intercepted arcs and divide by 2. Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and 37. CHALLENGE The circles below are concentric. ∠BCA are inscribed angles, by Theorem 11.6, m What is x? (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC).

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. SOLUTION: Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

Use the arcs intercepted on the smaller circle to find SOLUTION: m∠A. a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and

Use the arcs intercepted on the larger circle and x degrees. Hence solving leads to m∠A to find the value of x. the answer.

SOLUTION: By Theorem 11.13, So, 50° = Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and Therefore, x (minor arc) = 130, ∠BCA are inscribed angles, by Theorem 11.6, m and y (major arc) = 360° – 130° or 230°. (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC). 41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed 39. CCSS ARGUMENTS In the figure, is a by the points of tangency. diameter and is a tangent. SOLUTION: a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

Sample answer: Use Theorem 11.14 to find each minor arc. SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to The sum of the angles of a triangle is 180, so m∠R = the answer. 180 – (50 + 60) or 70.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your Therefore, the measures of the three minor arcs are reasoning. 130, 120, and 110. SOLUTION: 42. What is the value of x if and Sample answer:

By Theorem 11.13, So, 50° = A 23° B 31° Therefore, x (minor arc) = 130, C 64° D 128° and y (major arc) = 360° – 130° or 230°. SOLUTION: 41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how By Theorem 11.14, . to find the measures of the three minor arcs formed Substitute. by the points of tangency. SOLUTION: Simplify.

Sample answer: Use Theorem 11.14 to find each minor arc.

So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) The sum of the angles of a triangle is 180, so m R = ∠ G (10, –6) 180 – (50 + 60) or 70. H (5, –3) J (1, 5) SOLUTION:

Therefore, the measures of the three minor arcs are Here, C is the midpoint of . 130, 120, and 110. Use the midpoint formula, , to find the coordinates of 42. What is the value of x if and C.

So, the correct choice is J.

A 23° 44. GRIDDED RESPONSE If m∠AED = 95 and B 31° what is m∠BAC? C 64° D 128° SOLUTION:

By Theorem 11.14, . Substitute. SOLUTION:

Since . Simplify. We know that vertical angles are congruent.

So, . Since , . We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute.

11-6 Secants, Tangents, and Angle Measures So, the correct choice is C. Since , .

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on 45. SAT/ACT If the circumference of the circle below circle C, and is a diameter. What are the is 16π units, what is the total area of the shaded coordinates of C? regions? F (2, 10) G (10, –6) H (5, –3) J (1, 5) 2 SOLUTION: A 64π units B 32π units2 Here, C is the midpoint of . 2 Use the midpoint formula, C 12π units 2 , to find the coordinates of D 8π units 2 C. D 2π units SOLUTION: First, use the circumference to find the radius of the circle.

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and

what is m∠BAC? The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

SOLUTION:

Since .

We know that vertical angles are congruent. The area of the shaded regions is 32π. So, . So, the correct choice is B. Since , . Find x. Assume that segments that appear to be We know that the sum of the measures of all interior tangent are tangent. angles of a triangle is 180.

Substitute. 46. SOLUTION: Since , . By Theorem 11.10, . So, is a right 45. SAT/ACT If the circumference of the circle below triangle. is 16π units, what is the total area of the shaded Use the Pythagorean Theorem. regions? Substitute.

eSolutions Manual - Powered by Cognero Page 13 2 A 64π units B 32π units2 2 C 12π units D 8π units2 2 D 2π units SOLUTION: 47. First, use the circumference to find the radius of the circle. SOLUTION: If two segments from the same exterior point are tangent to a circle, then they are congruent.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. 48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x.

The area of the shaded regions is 32π. So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent. 49. PROOF Write a two-column proof. Given: is a semicircle;

46. Prove: SOLUTION: By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem.

Substitute. SOLUTION:

Given: is a semicircle.

Prove:

47. Proof: SOLUTION: Statements (Reasons) If two segments from the same exterior point are tangent to a circle, then they are congruent. 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.) 48. 7. (Def. of SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem. COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a Solve for x. degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION:

Use the Distance Formula to find the length of each side. 49. PROOF Write a two-column proof. Given: is a semicircle;

Prove:

Since , triangle BCD is a right triangle. SOLUTION: So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to

Given: is a semicircle. Write an equation using the cosine ratio.

Prove:

Use a calculator.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) Proof: Statements (Reasons) SOLUTION: 1. MHT is a semicircle; (Given) Use the Distance Formula to find the length of each 2. is a right angle. (If an inscribed angle side. intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.)

Since , triangle BCD is a right

triangle. Therefore, the solution is –9, –4. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to 2 Write an equation using the cosine ratio. 53. x – 6x = –9

SOLUTION:

Use a calculator.

Therefore, the solution is 3.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2 2), and Z(7, –2) 54. 3x + 15x = 0 SOLUTION: SOLUTION: Use the Distance Formula to find the length of each side.

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION: In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio.

Therefore, the solution is –7, 4.

If 56. x2 + 12x + 36 = 0 Use a calculator. SOLUTION:

Solve each equation. 52. x2 + 13x = –36 Therefore, the solution is –6. SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

Find each measure. Assume that segments that appear to be tangent are tangent. 1. 4.

SOLUTION: SOLUTION:

5. 2.

SOLUTION: SOLUTION:

So, the measure of arc TS is 144.

3. So, the measure of arc QTS is 248.

SOLUTION: SOLUTION:

So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several 4. barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

SOLUTION: Let x be the measure of the angle the ramp makes 5. with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION: Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. So, the measure of arc QTS is 248. 8.

SOLUTION: SOLUTION:

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15. ∠JMK and ∠HMJ form a linear pair.

Find each measure. Assume that segments that appear to be tangent are tangent. 8. 11.

SOLUTION: SOLUTION:

Therefore, the measure of arc RQ is 28. 9. 12.

SOLUTION: SOLUTION:

10. 13.

SOLUTION: SOLUTION:

∠JMK and ∠HMJ form a linear pair. So, the measure of arc PM is 144.

14.

11.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints. SOLUTION:

Therefore, the measure of arc RQ is 28.

12. 15.

SOLUTION: SOLUTION:

By Theorem 10.13, .

13.

16.

SOLUTION:

So, the measure of arc PM is 144. SOLUTION:

14. By Theorem 11.13, .

Solve for .

We know that . Substitute.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints. 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

15.

a. b. SOLUTION: SOLUTION: a. By Theorem 11.13, . By Theorem 10.13, . Substitute.

Simplify.

b. By Theorem 11.14,

. 16. Substitute.

Simplify.

SOLUTION: CCSS STRUCTURE Find each measure. 18. By Theorem 11.13, .

Solve for .

We know that . Substitute.

SOLUTION:

By Theorem 11.14, . 17. SPORTS The multi-sport field shown includes a Substitute. softball field and a soccer field. If , find each measure.

Simplify.

19. a. b. SOLUTION:

a. By Theorem 11.13, . Substitute.

SOLUTION: Simplify. By Theorem 11.14, . b. By Theorem 11.14, Substitute. . Substitute. Simplify.

Simplify.

CCSS STRUCTURE Find each measure. 20. m(arc JNM) 18.

SOLUTION:

By Theorem 11.14, . SOLUTION: Substitute.

Simplify. So, the measure of arc JNM is 205.

21.

19.

SOLUTION:

By Theorem 11.14, .

SOLUTION: Substitute.

By Theorem 11.14, . Simplify.

Substitute.

Simplify. 22.

20. m(arc JNM) SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

SOLUTION: 23.

So, the measure of arc JNM is 205. SOLUTION:

21. By Theorem 11.14, . Substitute.

Solve for .

SOLUTION: 24. JEWELRY In the circular necklace shown, A and B By Theorem 11.14, . are tangent points. If x = 260, what is y? Substitute.

Simplify.

22.

SOLUTION:

By Theorem 11.14, .

Substitute. SOLUTION:

By Theorem 11.14, . Simplify. Substitute.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

23.

SOLUTION: SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem By Theorem 11.14, . 11.14 to find the value of x. Substitute.

Solve for . Therefore, the measure of the planet’s arc that is visible to the satellite is 168. ALGEBRA Find the value of x.

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y? 26. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

27. 25. SPACE A satellite orbits above Earth’s equator. SOLUTION: Find x, the measure of the planet’s arc, that is visible

to the satellite. By Theorem 11.14, . Solve for .

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

28. Therefore, the measure of the planet’s arc that is visible to the satellite is 168. SOLUTION:

ALGEBRA Find the value of x. By Theorem 11.14, . Solve for .

26. SOLUTION:

By Theorem 11.14, . Solve for .

27. SOLUTION:

By Theorem 11.14, . Solve for . SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

28. b. Let x be the measure of the camera’s viewing SOLUTION: angle.

By Theorem 11.14, . By Theorem 11.14, . Solve for . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

SOLUTION: 4. (Substitution) a. Let x be the measure of the carousel that appears

in the shot. 5. (Subtraction Prop.)

By Theorem 11.14, . 6. (Distributive Prop.) Solve for .

31. Case 2 Given: tangent and secant

Prove: b. Let x be the measure of the camera’s viewing angle.

4. (Substitution)

5. (Subtraction Prop.)

4. (Substitution) SOLUTION: 5. (Subtraction Prop.) Statements (Reasons) 1. and are tangents to the circle. (Given) 6. (Distributive Prop.) 2. (The meas.

31. Case 2 of an secant-tangent the measure of its Given: tangent and secant intercepted arc.) Prove: 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.)

SOLUTION: 6. (Distributive Prop.) Statements (Reasons) 1. is a tangent to the circle and is a secant to 33. PROOF Write a paragraph proof of Theorem 11.13. the circle. (Given)

2. , ( The

meas. of an inscribed the measure of its

intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.) a. Given: is a tangent of . is a secant of .

6. (Distributive Prop.) ∠CAE is acute. Prove: m∠CAE = m(arc CA) 32. Case 3 b. Prove that if ∠CAB is obtuse, m∠CAB = m

Given: tangent and (arc CDA)

Prove: SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + SOLUTION: m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). Statements (Reasons) By substitution, 90 = m∠FAC + m∠CAE and 180 = 1. and are tangents to the circle. (Given) m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE 2. (The meas. = m(arc FC) + m(arc CA) by substitution. of an secant-tangent the measure of its m∠FAC = m(arc FC) since ∠FAC is inscribed, so intercepted arc.) substitution yields m(arc FC) + m∠CAE = m 3. (Exterior (arc FC) + m(arc CA). By Subtraction Prop., Theorem) m∠CAE = m(arc CA). 4. (Substitution) b. Proof: Using the Angle and Arc Addition 5. (Subtraction Prop.) Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since 6. (Distributive Prop.) and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a 33. PROOF Write a paragraph proof of Theorem 11.13. measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m∠CAB = m(arc

CF) + m(arc CDA) – m(arc CF). Then by subtraction, m∠CAB = m(arc CDA). a. Given: is a tangent of . is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA)

b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA) 34. WALLPAPER The design shown is an example of SOLUTION: optical wallpaper. is a diameter of If m∠A

a. Proof: By Theorem 11.10, . So, ∠FAE = 26 and = 67, what is is a right ∠ with measure of 90 and arc FCA is a Refer to the image on page 766. semicircle with measure of 180. Since ∠CAE is SOLUTION: acute, C is in the interior of FAE, so by the Angle ∠ First, find the measure of arc BD. and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + Since is a diameter, arc BDE is a semicircle and has a measure of 180. (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution.

m∠FAC = m(arc FC) since ∠FAC is inscribed, so 35. MULTIPLE REPRESENTATIONS In this substitution yields m(arc FC) + m∠CAE = m problem, you will explore the relationship between (arc FC) + m(arc CA). By Subtraction Prop., Theorems 11.12 and 11.6. a. GEOMETRIC Copy the figure shown. Then m∠CAE = m(arc CA). draw three successive figures in which the position of point D moves closer to point C, but points A, B, b. Proof: Using the Angle and Arc Addition and C remain fixed. Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since b. TABULAR Estimate the measure of for and is a diameter, ∠FAB is a right angle with a each successive circle, recording the measures of measure of 90 and arc FDA is a semicircle with a and in a table. Then calculate and record measure of 180. By substitution, m∠CAB = m∠CAF the value of x for each circle. + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by c. VERBAL Describe the relationship between substitution, m∠CAB = m(arc CF) + 90. Using the and the value of x as approaches zero. Division and Subtraction Properties on the Arc What type of angle does ∠AEB become when

Addition equation yields m(arc CDA) – m(arc

CF) = 90. By substituting for 90, m CAB = m(arc ∠ d. ANALYTICAL Write an algebraic proof to CF) + m(arc CDA) – m(arc CF). Then by show the relationship between Theorems 11.12 and 11.6 described in part c. subtraction, m∠CAB = m(arc CDA).

First, find the measure of arc BD. b.

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

c. As the measure of gets closer to 0, the MULTIPLE REPRESENTATIONS 35. In this measure of x approaches half of problem, you will explore the relationship between

Theorems 11.12 and 11.6. becomes an inscribed angle.

a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position d. of point D moves closer to point C, but points A, B, and C remain fixed. 36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a b. TABULAR Estimate the measure of for tangent that intersect outside a circle. each successive circle, recording the measures of SOLUTION: and in a table. Then calculate and record Find the difference of the two intercepted arcs and the value of x for each circle. divide by 2.

c. VERBAL Describe the relationship between 37. CHALLENGE The circles below are concentric. What is x? and the value of x as approaches zero. What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c. SOLUTION:

SOLUTION: a. Redraw the circle with points A, B, and C in the Use the arcs intercepted on the smaller circle to find same place but place point D closer to C each time. m∠A. Then draw chords and .

b. Use the arcs intercepted on the larger circle and m∠A to find the value of x.

38. REASONS Isosceles is inscribed in c. As the measure of gets closer to 0, the What can you conclude about and measure of x approaches half of Explain. becomes an inscribed angle.

36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: SOLUTION: Sample answer: m∠BAC = m∠BCA Find the difference of the two intercepted arcs and because the triangle is isosceles. Since ∠BAC and divide by 2. ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. 37. CHALLENGE The circles below are concentric. So, m(arc AB) = m(arc BC). What is x?

By Theorem 11.13, So, 50° =

Sample answer: Use Theorem 11.14 to find each minor arc.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to Therefore, the measures of the three minor arcs are

measure the angle that is formed. Find the measures 130, 120, and 110. of the minor and major arcs formed. Explain your reasoning. 42. What is the value of x if and

SOLUTION: Sample answer:

A 23° B 31° By Theorem 11.13, So, 50° = C 64° D 128° Therefore, x (minor arc) = 130, SOLUTION:

and y (major arc) = 360° – 130° or 230°. By Theorem 11.14, . 41. WRITING IN MATH A circle is inscribed within Substitute. If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. Simplify. SOLUTION:

Sample answer: Use Theorem 11.14 to find each minor arc. So, the correct choice is C.

Therefore, the measures of the three minor arcs are , to find the coordinates of 130, 120, and 110. C.

42. What is the value of x if and

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC? A 23° B 31° C 64° D 128° SOLUTION: SOLUTION: By Theorem 11.14, .

Substitute. Since . We know that vertical angles are congruent. So, . Simplify. Since , . We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute.

Since , .

SAT/ACT So, the correct choice is C. 45. If the circumference of the circle below is 16π units, what is the total area of the shaded 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on regions? circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) 2 H (5, –3) A 64π units J (1, 5) B 32π units2 2 SOLUTION: C 12π units 2 Here, C is the midpoint of . D 8π units 2 Use the midpoint formula, D 2π units , to find the coordinates of SOLUTION: C. First, use the circumference to find the radius of the circle.

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and The shaded regions of the circle comprise half of the what is m∠BAC? circle, so its area is half the area of the circle.

SOLUTION: The area of the shaded regions is 32π. Since . So, the correct choice is B. We know that vertical angles are congruent. Find x. Assume that segments that appear to be So, . tangent are tangent. Since , . We know that the sum of the measures of all interior angles of a triangle is 180. 46.

Substitute. SOLUTION:

By Theorem 11.10, . So, is a right triangle. Since , . Use the Pythagorean Theorem.

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded Substitute. regions?

circle.

The shaded regions of the circle comprise half of the 48. circle, so its area is half the area of the circle. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x.

11-6The Secants, area of Tangents, the shaded and regions Angle is 32Measuresπ. So, the correct choice is B.

Find x. Assume that segments that appear to be 49. PROOF Write a two-column proof. tangent are tangent. Given: is a semicircle;

Prove:

46. SOLUTION: By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem. SOLUTION:

Given: is a semicircle. Substitute.

Prove:

Proof: 47. Statements (Reasons) 1. MHT is a semicircle; (Given) SOLUTION: 2. is a right angle. (If an inscribed angle If two segments from the same exterior point are intercepts a semicircle, the angle is a right angle.) tangent to a circle, then they are congruent. 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.)

7. (Def. of

48. COORDINATE GEOMETRY Find the SOLUTION: measure of each angle to the nearest tenth of a Use Theorem 11.10 and the Pythagorean Theorem. degree by using the Distance Formula and an inverse trigonometric ratio. Solve for x. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each side.

49. PROOF Write a two-column proof. Given: is a semicircle;

Prove: eSolutions Manual - Powered by Cognero Page 14 Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio. SOLUTION: Given: is a semicircle.

If Prove: Use a calculator.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION:

Proof: Use the Distance Formula to find the length of each Statements (Reasons) side. 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.) In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite 7. (Def. of Write an equation using the sine ratio. COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. If 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) Use a calculator.

SOLUTION: Use the Distance Formula to find the length of each Solve each equation. 2 side. 52. x + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4. Since , triangle BCD is a right triangle. 2 So, CD is the length of the hypotenuse and BC is the 53. x – 6x = –9

length of the leg adjacent to SOLUTION: Write an equation using the cosine ratio.

Use a calculator. Therefore, the solution is 3.

2 55. 28 = x + 3x SOLUTION:

In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio. Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0

If SOLUTION:

Use a calculator.

Solve each equation. Therefore, the solution is –6. 52. x2 + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

Find each measure. Assume that segments that So, the measure of arc TS is 144. appear to be tangent are tangent. 3. 1.

SOLUTION: SOLUTION:

2. 4.

SOLUTION: SOLUTION:

So, the measure of arc TS is 144.

3. 5.

SOLUTION: SOLUTION:

So, the measure of arc QTS is 248. 4. 6.

SOLUTION: SOLUTION:

SOLUTION:

So, the measure of arc QTS is 248. SOLUTION: Let x be the measure of the angle the ramp makes 6. with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195.

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

So, the measure of arc LP is 150. Find each measure. Assume that segments that appear to be tangent are tangent. 7. STUNTS A ramp is attached to the first of several 8. barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

SOLUTION:

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 8. 10.

SOLUTION: SOLUTION:

and ∠JMK ∠HMJ form a linear pair.

11.

SOLUTION:

10. Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

SOLUTION: ∠JMK and ∠HMJ form a linear pair.

11. 13.

SOLUTION: SOLUTION:

So, the measure of arc PM is 144. Therefore, the measure of arc RQ is 28. 14. 12.

SOLUTION: Arc BD and arc BCD are a minor and major arc that SOLUTION: share the same endpoints.

13. 15.

SOLUTION:

So, the measure of arc PM is 144. By Theorem 10.13, .

14.

16. SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

SOLUTION:

By Theorem 11.13, . 15. Solve for .

We know that . Substitute.

SOLUTION: 17. SPORTS The multi-sport field shown includes a By Theorem 10.13, . softball field and a soccer field. If , find each measure.

16.

a. b. SOLUTION:

a. By Theorem 11.13, . SOLUTION: Substitute.

By Theorem 11.13, .

Solve for . Simplify.

b. By Theorem 11.14, We know that . Substitute. . Substitute.

17. SPORTS The multi-sport field shown includes a Simplify. softball field and a soccer field. If , find each measure.

CCSS STRUCTURE Find each measure. 18.

a. b. SOLUTION: SOLUTION: By Theorem 11.14, . a. By Theorem 11.13, . Substitute. Substitute.

Simplify. Simplify.

b. By Theorem 11.14,

. Substitute.

19. Simplify.

CCSS STRUCTURE Find each measure. 18. SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify.

By Theorem 11.14, . Substitute.

Simplify. 20. m(arc JNM)

19.

SOLUTION:

SOLUTION: So, the measure of arc JNM is 205. By Theorem 11.14, .

Substitute. 21.

Simplify.

SOLUTION: 20. m(arc JNM) By Theorem 11.14, . Substitute.

Simplify.

22.

SOLUTION:

SOLUTION: So, the measure of arc JNM is 205. By Theorem 11.14, . Substitute. 21.

Simplify.

23. SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify. By Theorem 11.14, . Substitute.

22.

Solve for .

SOLUTION: 24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y? By Theorem 11.14, . Substitute.

Simplify.

23. SOLUTION:

By Theorem 11.14, . Substitute.

SOLUTION: Simplify. By Theorem 11.14, . Substitute.

25. SPACE A satellite orbits above Earth’s equator.

Solve for . Find x, the measure of the planet’s arc, that is visible to the satellite.

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is SOLUTION: visible to the satellite is 168.

By Theorem 11.14, . ALGEBRA Find the value of x. Substitute.

Simplify. 26. SOLUTION:

By Theorem 11.14, . Solve for . 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

27.

Therefore, the measure of the planet’s arc that is SOLUTION: visible to the satellite is 168. By Theorem 11.14, . ALGEBRA Find the value of x. Solve for .

26. SOLUTION:

By Theorem 11.14, . Solve for .

28. SOLUTION:

By Theorem 11.14, . Solve for .

27. SOLUTION:

28. SOLUTION:

By Theorem 11.14, .

Solve for . SOLUTION:

a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

29. PHOTOGRAPHY A photographer frames a b. Let x be the measure of the camera’s viewing carousel in his camera shot as shown so that the lines angle. of sight form tangents to the carousel.

a. If the camera’s viewing angle is , what is the By Theorem 11.14, . arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in Solve for . the photograph, what viewing angle should be used?

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove: SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for . SOLUTION: Statements (Reasons) 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its b. Let x be the measure of the camera’s viewing angle. intercepted arc.) 3. (Exterior By Theorem 11.14, . Theorem) . Solve for 4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

CCSS ARGUMENTS For each case of 31. Case 2 Theorem 11.14, write a two-column proof. 30. Case 1 Given: tangent and secant

Given: secants and Prove:

Prove:

SOLUTION: SOLUTION: Statements (Reasons) Statements (Reasons) 1. is a tangent to the circle and is a secant to 1. and are secants to the circle. (Given) the circle. (Given) 2. , (The 2. , ( The

measure of an inscribed the measure of its meas. of an inscribed the measure of its intercepted arc.) intercepted arc.) 3. (Exterior 3. (Exterior Theorem) Theorem)

4. (Substitution) 4. (Substitution)

5. (Subtraction Prop.) 5. (Subtraction Prop.)

6. (Distributive Prop.) 6. (Distributive Prop.)

32. Case 3 31. Case 2

Given: tangent and Given: tangent and secant Prove: Prove:

SOLUTION: SOLUTION: Statements (Reasons) Statements (Reasons) 1. is a tangent to the circle and is a secant to 1. and are tangents to the circle. (Given) the circle. (Given) 2. , ( The 2. (The meas.

meas. of an inscribed the measure of its of an secant-tangent the measure of its

intercepted arc.) intercepted arc.) 3. (Exterior 3. (Exterior

Theorem) Theorem)

4. (Substitution) 4. (Substitution)

5. (Subtraction Prop.) 5. (Subtraction Prop.)

6. (Distributive Prop.) 6. (Distributive Prop.)

32. Case 3 33. PROOF Write a paragraph proof of Theorem 11.13.

Given: tangent and

Prove:

SOLUTION: a. Given: is a tangent of . Statements (Reasons) is a secant of . 1. and are tangents to the circle. (Given) ∠CAE is acute. 2. (The meas. Prove: m∠CAE = m(arc CA) b. Prove that if ∠CAB is obtuse, m∠CAB = m of an secant-tangent the measure of its (arc CDA) intercepted arc.) 3. (Exterior Theorem) SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE 4. (Substitution) is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is 5. (Subtraction Prop.) acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + 6. (Distributive Prop.) m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = 33. PROOF Write a paragraph proof of Theorem 11.13. m(arc FC) + m(arc CA). So, 90 = m(arc FC) +

(arc CA) by Division Prop., and m∠FAC + m∠CAE

= m(arc FC) + m(arc CA) by substitution. m∠FAC = m(arc FC) since ∠FAC is inscribed, so

substitution yields m(arc FC) + m∠CAE = m

(arc FC) + m(arc CA). By Subtraction Prop., m∠CAE = m(arc CA).

b. Proof: Using the Angle and Arc Addition

a. Given: is a tangent of . Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar is a secant of . CDA) = m(arc CF) + m(arc FDA). Since ∠CAE is acute. and is a diameter, ∠FAB is a right angle with a Prove: m∠CAE = m(arc CA) measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF b. Prove that if ∠CAB is obtuse, m∠CAB = m + 90 and m(arc CDA) = m(arc CF) + 180. Since (arc CDA) ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the SOLUTION: Division and Subtraction Properties on the Arc a. Proof: By Theorem 11.10, . So, FAE ∠ Addition equation yields m(arc CDA) – m(arc is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is CF) = 90. By substituting for 90, m∠CAB = m(arc acute, C is in the interior of ∠FAE, so by the Angle CF) + m(arc CDA) – m(arc CF). Then by and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). subtraction, m∠CAB = m(arc CDA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) +

(arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution. m∠FAC = m(arc FC) since ∠FAC is inscribed, so

substitution yields m(arc FC) + m∠CAE = m 34. WALLPAPER The design shown is an example of (arc FC) + m(arc CA). By Subtraction Prop., optical wallpaper. is a diameter of If m∠A m∠CAE = m(arc CA). = 26 and = 67, what is Refer to the image on page 766. b. Proof: Using the Angle and Arc Addition SOLUTION: Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar First, find the measure of arc BD. CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF Since is a diameter, arc BDE is a semicircle and + 90 and m(arc CDA) = m(arc CF) + 180. Since has a measure of 180. ∠CAF is inscribed, m∠CAF = m(arc CF) and by

substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc 35. MULTIPLE REPRESENTATIONS In this Addition equation yields m(arc CDA) – m(arc problem, you will explore the relationship between Theorems 11.12 and 11.6. CF) = 90. By substituting for 90, m∠CAB = m(arc a. GEOMETRIC Copy the figure shown. Then CF) + m(arc CDA) – m(arc CF). Then by draw three successive figures in which the position of point D moves closer to point C, but points A, B, subtraction, m∠CAB = m(arc CDA). and C remain fixed.

b. TABULAR Estimate the measure of for each successive circle, recording the measures of and in a table. Then calculate and record the value of x for each circle.

c. VERBAL Describe the relationship between 34. WALLPAPER The design shown is an example of and the value of x as approaches zero. optical wallpaper. is a diameter of If m∠A What type of angle does ∠AEB become when = 26 and = 67, what is Refer to the image on page 766. SOLUTION: d. ANALYTICAL Write an algebraic proof to First, find the measure of arc BD. show the relationship between Theorems 11.12 and 11.6 described in part c.

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

SOLUTION: 35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between a. Redraw the circle with points A, B, and C in the Theorems 11.12 and 11.6. same place but place point D closer to C each time. a. GEOMETRIC Copy the figure shown. Then Then draw chords and . draw three successive figures in which the position of point D moves closer to point C, but points A, B, and C remain fixed.

b. b. TABULAR Estimate the measure of for each successive circle, recording the measures of and in a table. Then calculate and record the value of x for each circle.

c. VERBAL Describe the relationship between

and the value of x as approaches zero. c. As the measure of gets closer to 0, the What type of angle does ∠AEB become when measure of x approaches half of becomes an inscribed angle.

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and d. 11.6 described in part c. 36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: Find the difference of the two intercepted arcs and divide by 2. SOLUTION: 37. CHALLENGE The circles below are concentric. a. Redraw the circle with points A, B, and C in the What is x? same place but place point D closer to C each time. Then draw chords and .

b. SOLUTION:

c. As the measure of gets closer to 0, the

measure of x approaches half of Use the arcs intercepted on the smaller circle to find becomes an inscribed angle. m∠A.

36. WRITING IN MATH Explain how to find the

measure of an angle formed by a secant and a Use the arcs intercepted on the larger circle and tangent that intersect outside a circle. m∠A to find the value of x. SOLUTION: Find the difference of the two intercepted arcs and divide by 2.

37. CHALLENGE The circles below are concentric. What is x? 38. REASONS Isosceles is inscribed in What can you conclude about and Explain.

SOLUTION: SOLUTION: Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC.

Use the arcs intercepted on the smaller circle to find So, m(arc AB) = m(arc BC). m∠A.

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. Use the arcs intercepted on the larger circle and b. If m∠G = 34, find the measures of minor arcs HJ m∠A to find the value of x. and KH. Explain.

38. REASONS Isosceles is inscribed in SOLUTION: What can you conclude about and a. ; for all values except when

Explain. at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer. SOLUTION: 40. OPEN ENDED Draw a circle and two tangents Sample answer: m∠BAC = m∠BCA that intersect outside the circle. Use a protractor to because the triangle is isosceles. Since ∠BAC and measure the angle that is formed. Find the measures ∠BCA are inscribed angles, by Theorem 11.6, m of the minor and major arcs formed. Explain your

(arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. reasoning. So, m(arc AB) = m(arc BC). SOLUTION: Sample answer:

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain.

b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain. By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°.

41. WRITING IN MATH A circle is inscribed within SOLUTION: If m∠P = 50 and m∠Q = 60, describe how a. ; for all values except when to find the measures of the three minor arcs formed at G, then . by the points of tangency. b. Because a diameter is SOLUTION: involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer.

40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures Sample answer: Use Theorem 11.14 to find each of the minor and major arcs formed. Explain your minor arc. reasoning. SOLUTION: Sample answer:

The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70. By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°. Therefore, the measures of the three minor arcs are 130, 120, and 110. 41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how 42. What is the value of x if and to find the measures of the three minor arcs formed by the points of tangency. SOLUTION:

A 23° B 31° C 64° D 128° Sample answer: Use Theorem 11.14 to find each minor arc. SOLUTION: By Theorem 11.14, . Substitute.

Simplify.

The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70.

Therefore, the measures of the three minor arcs are 130, 120, and 110.

42. What is the value of x if and So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) A 23° J (1, 5) B 31° C 64° SOLUTION: D 128° Here, C is the midpoint of . Use the midpoint formula, SOLUTION: , to find the coordinates of By Theorem 11.14, . C. Substitute.

Simplify. So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

So, the correct choice is C. SOLUTION: ALGEBRA 43. Points A(–4, 8) and B(6, 2) are both on Since . circle C, and is a diameter. What are the coordinates of C? We know that vertical angles are congruent. F (2, 10) So, . G (10, –6) Since , H (5, –3) . J (1, 5) We know that the sum of the measures of all interior angles of a triangle is 180. SOLUTION: Here, C is the midpoint of . Substitute. Use the midpoint formula, , to find the coordinates of Since , . C. 45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and 2 what is m∠BAC? A 64π units B 32π units2 2 C 12π units D 8π units2 2 D 2π units SOLUTION: SOLUTION: First, use the circumference to find the radius of the Since . circle. We know that vertical angles are congruent. So, . Since , . We know that the sum of the measures of all interior angles of a triangle is 180. The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions? The area of the shaded regions is 32π. So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent.

2 A 64π units B 32π units2 2 C 12π units 46. D 2 8π units SOLUTION: 2 D 2π units By Theorem 11.10, . So, is a right SOLUTION: triangle.

First, use the circumference to find the radius of the Use the Pythagorean Theorem. circle. Substitute.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

47. SOLUTION: If two segments from the same exterior point are The area of the shaded regions is 32π. tangent to a circle, then they are congruent. So, the correct choice is B. Find x. Assume that segments that appear to be tangent are tangent.

48. 46. SOLUTION: SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem. By Theorem 11.10, . So, is a right triangle. Solve for x. Use the Pythagorean Theorem.

Substitute.

49. PROOF Write a two-column proof.

Given: is a semicircle;

Prove:

47. SOLUTION: If two segments from the same exterior point are SOLUTION: tangent to a circle, then they are congruent. Given: is a semicircle.

Prove:

48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x. Proof:

Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 49. PROOF Write a two-column proof. 5. (Reflexive Prop.) Given: is a semicircle; 6. (AA Sim.)

Prove: 7. (Def. of

COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio.

50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – SOLUTION: 5), and D(–1, 2) Given: is a semicircle. SOLUTION: Use the Distance Formula to find the length of each side. Prove:

Since , triangle BCD is a right

Proof: triangle.

Statements (Reasons) So, CD is the length of the hypotenuse and BC is the

1. MHT is a semicircle; (Given) length of the leg adjacent to 2. is a right angle. (If an inscribed angle Write an equation using the cosine ratio. intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) If 6. (AA Sim.)

Use a calculator. 11-67. Secants, Tangents, (Def. of and Angle Measures

COORDINATE GEOMETRY Find the 51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – measure of each angle to the nearest tenth of a 2), and Z(7, –2) degree by using the Distance Formula and an inverse trigonometric ratio. SOLUTION: 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – Use the Distance Formula to find the length of each 5), and D(–1, 2) side. SOLUTION: Use the Distance Formula to find the length of each side.

In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio. Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to If Write an equation using the cosine ratio.

Use a calculator.

Solve each equation. If 52. x2 + 13x = –36 Use a calculator. SOLUTION:

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Therefore, the solution is –9, –4. Use the Distance Formula to find the length of each side. 2 53. x – 6x = –9

SOLUTION:

In right triangle XYZ, ZX is the length of the Therefore, the solution is 3. hypotenuse and YZ is the length of the leg opposite 2 54. 3x + 15x = 0 Write an equation using the sine ratio. SOLUTION:

If eSolutions Manual - Powered by Cognero Page 15 Use a calculator. Therefore, the solution is –5, 0.

2 Solve each equation. 55. 28 = x + 3x 52. x2 + 13x = –36 SOLUTION: SOLUTION:

Therefore, the solution is –9, –4. Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 2 53. x – 6x = –9 SOLUTION: SOLUTION:

Therefore, the solution is –6.

Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6. Find each measure. Assume that segments that appear to be tangent are tangent. 1.

SOLUTION:

So, the measure of arc TS is 144.

SOLUTION:

So, the measure of arc QTS is 248.

SOLUTION:

So, the measure of arc LP is 150.

7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground?

Therefore, the measure of the angle the ramp makes with the ground is 15.

Find each measure. Assume that segments that appear to be tangent are tangent. 8.

SOLUTION:

10.

SOLUTION:

∠JMK and ∠HMJ form a linear pair.

11.

SOLUTION:

Therefore, the measure of arc RQ is 28.

12.

SOLUTION:

13.

SOLUTION:

So, the measure of arc PM is 144.

14.

SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints.

15.

SOLUTION:

By Theorem 10.13, .

16.

SOLUTION:

By Theorem 11.13, .

Solve for .

We know that . Substitute.

17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If , find each measure.

a. b. SOLUTION:

a. By Theorem 11.13, . Substitute.

Simplify.

b. By Theorem 11.14,

. Substitute.

Simplify.

CCSS STRUCTURE Find each measure. 18.

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

19.

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

20. m(arc JNM)

SOLUTION:

So, the measure of arc JNM is 205.

21.

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

22.

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

23.

SOLUTION:

By Theorem 11.14, . Substitute.

Solve for .

24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y?

SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite.

SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 11.14 to find the value of x.

Therefore, the measure of the planet’s arc that is visible to the satellite is 168.

ALGEBRA Find the value of x.

26. SOLUTION:

By Theorem 11.14, . Solve for .

27. SOLUTION:

By Theorem 11.14, . Solve for .

28. SOLUTION:

By Theorem 11.14, . Solve for .

SOLUTION: a. Let x be the measure of the carousel that appears in the shot.

By Theorem 11.14, . Solve for .

b. Let x be the measure of the camera’s viewing angle.

By Theorem 11.14, . Solve for .

CCSS ARGUMENTS For each case of Theorem 11.14, write a two-column proof. 30. Case 1 Given: secants and

Prove:

SOLUTION: Statements (Reasons) 1. and are secants to the circle. (Given) 2. , (The

measure of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

31. Case 2 Given: tangent and secant

Prove:

SOLUTION: Statements (Reasons) 1. is a tangent to the circle and is a secant to the circle. (Given) 2. , ( The

meas. of an inscribed the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

32. Case 3 Given: tangent and

Prove:

SOLUTION: Statements (Reasons) 1. and are tangents to the circle. (Given)

2. (The meas.

of an secant-tangent the measure of its intercepted arc.) 3. (Exterior Theorem)

4. (Substitution)

5. (Subtraction Prop.)

6. (Distributive Prop.)

33. PROOF Write a paragraph proof of Theorem 11.13.

a. Given: is a tangent of . is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA) b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA)

SOLUTION: a. Proof: By Theorem 11.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution. m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m (arc FC) + m(arc CA). By Subtraction Prop., m∠CAE = m(arc CA).

34. WALLPAPER The design shown is an example of optical wallpaper. is a diameter of If m∠A = 26 and = 67, what is Refer to the image on page 766. SOLUTION: First, find the measure of arc BD.

Since is a diameter, arc BDE is a semicircle and has a measure of 180.

b. TABULAR Estimate the measure of for each successive circle, recording the measures of and in a table. Then calculate and record the value of x for each circle.

c. VERBAL Describe the relationship between and the value of x as approaches zero. What type of angle does ∠AEB become when

d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 11.12 and 11.6 described in part c.

SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and .

c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle.

36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: Find the difference of the two intercepted arcs and divide by 2.

37. CHALLENGE The circles below are concentric. What is x?

SOLUTION:

Use the arcs intercepted on the smaller circle to find m∠A.

Use the arcs intercepted on the larger circle and m∠A to find the value of x.

38. REASONS Isosceles is inscribed in What can you conclude about and Explain.

SOLUTION: Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by Theorem 11.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC).

39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain.

SOLUTION: a. ; for all values except when at G, then . b. Because a diameter is involved the intercepted arcs measure (180 – x) and x degrees. Hence solving leads to the answer.

By Theorem 11.13, So, 50° =

Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°.

41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. SOLUTION:

Sample answer: Use Theorem 11.14 to find each minor arc.

The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70.

Therefore, the measures of the three minor arcs are 130, 120, and 110.

42. What is the value of x if and

A 23° B 31° C 64° D 128° SOLUTION:

By Theorem 11.14, . Substitute.

Simplify.

So, the correct choice is C.

43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) J (1, 5) SOLUTION: Here, C is the midpoint of . Use the midpoint formula, , to find the coordinates of C.

So, the correct choice is J.

44. GRIDDED RESPONSE If m∠AED = 95 and what is m∠BAC?

SOLUTION:

Since . We know that vertical angles are congruent. So, . Since , . We know that the sum of the measures of all interior angles of a triangle is 180.

Substitute.

Since , .

45. SAT/ACT If the circumference of the circle below is 16π units, what is the total area of the shaded regions?

2 A 64π units B 32π units2 2 C 12π units D 8π units2 2 D 2π units SOLUTION: First, use the circumference to find the radius of the circle.

The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle.

The area of the shaded regions is 32π. So, the correct choice is B.

Find x. Assume that segments that appear to be tangent are tangent.

46. SOLUTION: By Theorem 11.10, . So, is a right triangle. Use the Pythagorean Theorem.

Substitute.

47. SOLUTION: If two segments from the same exterior point are tangent to a circle, then they are congruent.

48. SOLUTION: Use Theorem 11.10 and the Pythagorean Theorem.

Solve for x.

49. PROOF Write a two-column proof. Given: is a semicircle;

Prove:

SOLUTION: Given: is a semicircle.

Prove:

7. (Def. of

COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 50. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each side.

Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio.

Use a calculator.

51. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Use the Distance Formula to find the length of each side.

In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite

Write an equation using the sine ratio.

Use a calculator.

Solve each equation. 52. x2 + 13x = –36 SOLUTION:

Therefore, the solution is –9, –4.

2 53. x – 6x = –9 SOLUTION:

11-6 Secants, Tangents, and Angle Measures Therefore, the solution is 3.

54. 3x2 + 15x = 0 SOLUTION:

Therefore, the solution is –5, 0.

2 55. 28 = x + 3x SOLUTION:

Therefore, the solution is –7, 4.

56. x2 + 12x + 36 = 0 SOLUTION:

Therefore, the solution is –6.