First Year of Algebra

Algebra 113

Supplementary Unit on Trigonometry

Written by:

Wally Dodge Pat Wadecki Kathy Anderson Becky Lee David Fickett Jean Hostetler

(modified) First Year Algebra 113 Supplemental Trigonometry Section #1

In this section you will learn the following: A) The definition of the 3 trigonometry ratios: sine (sin), cosine (cos), and tangent (tan) B) How to use your calculator to find trigonometric ratios C) How to use trigonometry to solve problems.

Consider the following right triangle.

hypotenuse leg opposite angle 

 leg adjacent to angle 

In mathematics we often use the Greek letter , pronounced “theta,” to indicate an angle in a triangle. The leg that forms one of the sides of  is called the leg adjacent to . The other leg is called the leg opposite from .

Example 1: Complete the chart for each triangle given below. III 8  I 5 3 15  4

12  5 13 II

Table #1 Triangle Leg Leg hypotenuse Ratio of Ratio of Ratio of opposite  adjacent  opposite leg adjacent leg opposite leg to hypotenuse to hypotenuse to adjacent leg I II III

2 Solution to example 1 from previous page:

Table #1 Ratio of Ratio of adjacent Ratio of opposite Triangle Leg Leg hypotenuse opposite leg to leg to leg to adjacent opposite  adjacent  hypotenuse hypotenuse leg I 3 4 5 3/5 = 0.6 4/5 = 0.8 3/4 = 0.75 II 5 12 13 5/13 ≈ 0.385 12/13 ≈ 0.923 5/12 ≈ 0.417 III 15 8 17* 15/17 ≈ 0.882 8/17 ≈ 0.471 15/8 = 1.875

*Note: We used the Pythagorean Theorem for triangle III, in order to find that the hypotenuse was 17.

Using a protractor to measure the angles marked  in each of the triangles labeled I, II, and III on the previous page, we found the information summarized in the table below.

Table #2 Triangle Measure of  in degrees I ≈ 36.9° II ≈ 22.6° III ≈ 61.9°

Example 2: Set your calculator into “Degree” mode and complete the following table by choosing the appropriate keys on your calculator.

Table #3 in sin() cos() tan() Triangle degrees I 36.9° II 22.6° III 61.9°

Solution to example 2 from above:

Table #3 in sin() cos() tan() Triangle degrees I 36.9° ≈ 0.600 ≈ 0.800 ≈ 0.751 II 22.6° ≈ 0.384 ≈ 0.923 ≈ 0.416 III 61.9° ≈ 0.882 ≈ 0.471 ≈ 1.873

If you compare the columns with the ratios in Table #1 with the sin(), cos(), and tan() columns in Table #3, you will notice something very interesting. The results are nearly identical. This not a coincidence: these ratios are the definitions of sin(), cos(), and tan(). B Definition: In a right triangle with an acute angle labeled , the trigonometric ratios sine(), cosine(), and tangent() are definedhypotenuse as follows: leg opposite angle 

 3 A leg adjacent to angle  C leg opposite angle  BC sine(), abbreviated sin() = , or sin(A) = hypotenuse AB

leg adjacent angle  AC cosine(), abbreviated cos() = , or cos(A) = hypotenuse AB

leg opposite angle  BC tangent(), abbreviated tan() = , or tan(A) = leg adjacent angle  AC

Example 3: Given the figure below, find sin(M ), cos(M), and tan(M) 12 P N

Solution to Example 3: By the Pythagorean Theorem, MP = 15 12 P N opposite leg PN 12 4 sin(M) =    = 0.8 hypotenuse PM 15 5 9 15 adjacent leg MN 9 3 cos(M) =    = 0.6 hypotenuse PM 15 5 M opposite leg PN 12 4 tan(M) =    ≈ 1.333 adjacent leg MN 9 3

Example 4: For the figure given below of a right triangle with an acute angle of 35°, do the following:

4 a) Using your knowledge of trigonometric ratios, write two equations, one involving x and the other involving y.

b) Solve the equations that your wrote in part (a).

35° 4 Solution to Example 4:

x 4 a) tan(35°) = and cos(35°) = 4 y x Note: if you use the trigonometric ratio for sine you get sin(35°) = which is an equation with both x and y y. It is a correct equation, but, since it has both variables in it, this equation isn’t useful to us. b) Using the trig keys on the calculator (make sure you’re still in “degree” mode!), we obtain:

x 4 tan(35°) = cos(35°) = 4 y x 4 0.7002075 ≈ .819152 ≈ 4 y 0.7002075*4 ≈ x .819152*y ≈ 4 4 0.7002075*4 ≈ x y ≈ .819152 2.80 ≈ x y ≈ 4.88

Practice Problems Using Trigonometry (solutions follow on pages 8 and 9)

1. Find the height of the flagpole given the information shown on the diagram.

2. An airplane is 50 miles from an airport and the angle of depression of the airport is 7°. How high is the airplaneairplane above the ground? 51° 7° 36 feet

5 50 miles

airport 3. A picture that is 4 feet wide is to be hung from a wire making an angle of 40° with the picture, as shown below. How long is the wire? wire

Nail●

40° 40°

4 feet

4. The Marina is directly across Springstead Lake from the Frontier Inn. To get from one to the other without getting your feet wet, you measure horizontally 4575 feet from the Frontier Inn to point A, and find that the measure of A is 25°. What is the direct distance, across the lake, from the Marina to the Frontier Inn?

Frontier Inn 4575 feet A 25°

Marina

6 SOLUTIONS to Section #1: Practice Problems Using Trigonometry

#1) Let x = the height of the flagpole, in feet. From the diagram, we have: x tan(51°) = 36 x = 36 * tan(51°) calculator gives x ≈ 44.46

So the height of the flagpole is approximately 44.46 feet.

#2) The angle of depression is the angle measured from the horizontal to the direction looking downward, as shown in the diagram. (Further explanation on the angle of depression is given on page 9.) Let x = the height of the airplane. From the diagram, we have: x sin(7°) = 50 x = 50 * sin(7°) calculator gives x ≈ 6.095

The airplane is approximately 6.095 miles above ground, which is a little over 32,000 feet. (This is a reasonable cruising altitude for a commercial airliner.)

#3) To use trigonometry, we need a right triangle, so we draw in the height of the triangle formed by the wire. This height divides the horizontal side of the picture into two equal pieces. Let x = one-half the length of the wire, as shown in the diagram below.

From this diagram (with the height drawn in), we have: x 40° 2 2 cos(40°) = x

x * cos(40°) = 2 2 x = ≈ 2.61 cos(40° ) 4 feet

Since x represents half the length of the wire, the wire is approximately 5.22 feet long.

7 More SOLUTIONS to Section #1: Practice Problems Using Trigonometry

#4) Let x = the distance, in feet, across the lake from the Marina to the Inn, as shown below.

From this diagram, we have: Frontier Inn 4575 feet A x 25° tan(25°) = 4575 x

x = 4575 * tan(25°) x ≈ 2133.36 Marina

The distance across the lake from the Marina to the Frontier Inn is approximately 2133 feet.

NOTE: For some applications of trigonometry, you need to know the meanings of angle of elevation and angle of depression.

A If an observer at point P looks upward toward an object at A, the angle the line of  sight PA makes with the horizontal  angle of elevation PH is called the angle of elevation. P H

If an observer at point P looks downward angle of depression toward an object at B, the angle the line of  sight PB makes with the horizontal  PH is called the angle of depression. B

(from: Rhoad, Milauskus, & Whipple: Geometry for Enjoyment and Challenge, New Edition. McDougal, Little, & Company: 1991)

In class Example: 1. Tom is looking up at his friend, Jen who is in a tree. He estimates his angle of elevation to be 40⁰ and is standing 20 feet away from the base of the tree. How high off the ground is Jen, to the nearest foot?

2. Jack is looking out the window at Jill. He estimates his angle of depression to be 30⁰. He know he’s about 45 feet above ground level. How far is Jill from the building?

8 Exercises for Trigonometry: Section I B 1) a) Label the sides of the triangle opposite, hypotenuse, and adjacent using A as the given angle.

b) Label the sides of the triangle opposite, hypotenuse, and adjacent using B as the given angle.

2) For the given triangles, with angle , do the following: i) Give the lengths of the side adjacent to , the side opposite , and the hypotenuse. ii) Give the values of sin(), cos(), and tan().

4 a) b)

3 13 5  

10 c) d) 6  45

12 

3) Find sin(A), cos(A), and tan(A) in terms of a, b, and c. a B

9 4) Using your knowledge of trigonometry, find the length of the missing side marked x on each figure.

x a) b) 15

° 10 ° x

c) d) ° x x ° 12 14

5) Find the value of x in each case.

a) b) c) 40° x 3 25° x 7 45° 4 x

6) Given a right triangle ABC with the right angle at C. a) Draw a possible picture of this right triangle. b) If sin(A) = 3/5, then cos(A) = c) If tan(A) = 2/3, then sin(A) = (info from previous parts no longer true) d) If cos(A) = 4/7, then tan(A) = (info from previous parts no longer true)

7) Round your answers to this problem to the nearest hundredth. (Use the triangle shown!)

60° 2 1

30° 3

Find: a) sin(30°) d) cos(30°) b) cos(60°) e) tan(30°) c) sin(60°) f) tan(60°)

10 8) Round your answers to the nearest hundredth. 5 5

45° 5 2 Find: a) sin(45°) b) cos(45°) c) tan(45°)

d) Why are sin(45°) and cos(45°) the same?

9) You are flying a kite on a string that is 150 feet in length. The angle that the kite string makes with the ground is 75°. Find the height of the kite above the ground.

10) You are parasailing behind a boat. You are 125 feet above the water, and the rope that you are holding (attached to the boat) makes an angle of 50° with the water. How long is the rope that you are holding?

11) To find the distance across Lake Springstead without getting his feet wet, Sam the surveyor measures the distance (3600 feet) and the angle (22°) shown in the diagram below. If B is a right angle, how far is it across Lake Springstead?

3600 feet A 22°

12) You are 60 feet due west from the front of the Eurostar train in Waterloo station, which is facing North/South. You calculate an angleB of 60° between the line from where you are standing to the front of the train and the line from where you are standing to the rear of the train. How long is the train?

11 13) You are at an elevation of 18,500 feet and you spot the peak of Mt. McKinley about one mile away along your line of sight. You can’t remember how high Mt. McKinley is, so you take out some of your surveying equipment to find the angle of elev. You find that the angle of elevation of the peak of Mt.McKinley, from where you are standing, is approximately 20.2°. How high is this mountain peak? Mt. McKinley

14) A guard standing on the lookout on top of a castle 250 feet high has spotted an army of knights approaching at an angle of depression of 2°. How far is this army from the castle?

15) Heimlich maneuvered his mountain bike straight up the side of a hill with a 16° grade. If the hill had an elevation (= height) of 1600 feet, how long was the trail up the hill?

16) Mario parked his car at a 45° angle with the curb, as shown in the diagram. If the car is 14 feet long and 6 feet wide, how far is the right rear corner of car 45° the car from curb? How far is the left front corner from the curb?

right rear corner

17) Mrs. Bluebird is on a mad hunt for a worm to give to her babies for breakfast. She is flying at an altitude of 34 feet when she spots a worm at an angle of depression of 19°. How far must she travel to catch breakfast for her babies?

18) Rectangle ABCD is inscribed in circle O. (That is, it just fits inside the circle, and it’s corners are labeled A, B, C, D as you walk around the circle.) If the radius of circle O is 19 and the measure of BDC is 30°, find the perimeter of triangle ABD.

19) A satellite hovers 300 km above the planet. The angle from the satellite to the horizon is 80°, as shown in the diagram. What is the radius of the planet?

(satellite) ● P

300 km

12 First Year Algebra 113 Supplemental Trigonometry Section #2

In this section you will learn the following: A) How to find an angle when given a trigonometric ratio B) How to use your calculator to find angles C) How to use trigonometry to do application problems.

In the last section you learned how to find lengths of sides in a right triangle when given one side and an acute angle in the triangle. In this section you will learn how to find an acute angle in a right triangle when given two sides of that triangle.

Example 1: Use your graphing calculator’s “table” tool to find an approximate value for the angle shown in each triangle below. I II  5 4 6  7 4 Solution For Triangle I: tan() = ≈ 0.5714; in “degree” mode, let y1 = tan(x) and set up a 7 table to start at 0° and move in steps of 1°, as shown below. To the nearest degree,  ≈ 30°.

5 For Triangle II: sin() = ≈ 0.8333; this time, let y1 = sin(x) and again set up a table to 6 start at 0° and move in steps of 1°, as shown above. To the nearest degree,  ≈ 56°.

Earlier this year, we discussed inverse operations. We learned, for example, that the inverse operation for dividing is to multiply, and that the inverse operation for square root is to square.

What we need is a function that would do the inverse operation of a trigonometric function. That 5 is, suppose that sin() = . What operation could we do to both sides of the equation in order 6 to find the value of ?

Fortunately, there is an inverse operation for sine, called the inverse sine function. This function is illustrated in the following example. 13 5 Example 2: Suppose sin() = . What is the size of the angle , in degrees? 6

Solution: The inverse of the sine function is the labeled “sin−1” on your calculator – it is located above the “sin” key. In order to find the angle  we do the following: 5 sin() = 6 5 sin−1(sin()) = sin−1( ) 6 5 Since sin−1 is the inverse operation for sine we obtain  = sin−1( ). We use a calculator to 6 5 find that sin−1( )≈ 56.44°. 6

Example 3: Nate is watching a ball game from the top of Comiskey Park. His seat is 130 feet above the playing field and his distance from the pitcher is 160 feet. At what angle does he need to tilt his head down in order to see the pitcher?

Solution: Below is a drawing to describe the situation

Nate 

130 ft 160 ft

pitcher

130 We see the following trigonometric ratio: sin() = . Using our inverse operation for 160 sine, we obtain 130 sin−1(sin()) = sin−1( ) 160 130  = sin−1( ) ≈ 54.34° 160

14 Although we have only used the inverse operation for sine, namely sin−1, there are also inverse functions for both the cosine and tangent. We formalize this in the following definition:

Definition: a) The inverse operation for sine, called the inverse sine function, is denoted by sin−1. b) The inverse operation for cosine, called the inverse cosine function, is denoted by cos−1. c) The inverse operation for tangent, called the inverse tangent function, is denoted by tan−1.

Practice Problems Using Inverse Trigonometric Functions (solutions follow on p. 18 and 19)

1. Find the measures of A and C in the right triangle given below:

12 A B

2. Find the value of x and the value of  in the figure below.

40°  9 12

3. If you are standing at the origin of a rectangular coordinate system, and are facing the positive x-axis, by what angle do you have to turn in order to look directly at the ordered pair (5, 15) ? y ● (5, 15)

15 SOLUTIONS to Section II: Practice Problems Using Inverse Trigonometric Functions

#1) 12 A B

C From the diagram: 5 12 tan(A) = tan(C) = 12 5 5 12 tan−1(tan(A)) = tan−1( ) tan−1(tan(C)) = tan−1( ) 12 5 5 12 A = tan−1( ) C = tan−1( ) 12 5 A ≈ 22.6° C ≈ 67.4°

Note: Once you found that A ≈ 22.6°, if you remember that all angles in a triangle add up to 180°, you can find C by subtraction: C = 180 − 90 − 22.6 = 67.4°.

40°  9 12

Using the right triangle at the left of the picture, we have: x tan(40°) = . Therefore 9∙tan(40°) = x, and thus x ≈ 7.5519 9 Now using this result and the right triangle at the right of the picture we obtain: x 7.5519 tan() = ≈ 12 12 7.5519 tan−1(tan()) = tan−1( ) 12  ≈ 32.18°

Note: Be careful about rounding error….. see how we used four decimals in our approximation of x when we used it to find . Better yet, use your calculator’s “ANS” key!

16 MORE solutions to Section II: Practice Problems Using Inverse Trigonometric Functions

#3) We first add some lines to the original diagram to obtain:

y (5, 15) ● 15 tan() = = 3 5 tan−1(tan()) = tan−1(3)

 ≈ 71.57°  x

NOTE The sum of the angles in a triangle is:

Isosceles Triangles:

Exercises for Trigonometry: Section II

1. Find the missing angle labeled A in each diagram.

A a) b) 5 3

3 3  c) d) 3 4 7

17 2. Find the values of x and y in the diagram below.

24 12 x

40° y

3. Find the values of x and y in the diagram at right. 6.5 (Both x and y are labeling angles.) 4 x y 4 4

4. An isosceles triangle ABC has two equal sides of 14 inches long, and the base is 8 inches long. Find the measures of A, C, and ABC (in degrees). B

5. You are standing at the origin of a rectangular coordinate system, facing out towards the positive x-axis. How many degrees do you rotate counter-clockwise to be facing directly at the point (2, 21) ?

6. From the right bank of the Seine, you spot the top of the 985 foot high Eiffel Tower, one mile in the distance. Find the angle of elevation to the top of the tower.

7. You are flying at an altitude of 3300 feet and spot the London Bridge, 5000 feet from where you are. Find the angle of depression from you to the bridge.

8. Dubious wanted to make a skateboard ramp by nailing ramp plywood over 4 stairs on his family’s backyard deck. Each stair had a run (depth) of 10 inches and a rise of 8 inches. What would be the angle of elevation of the ramp?

18 9. In right ABC with right angle at C, the length AC = 4.32 cm. and the length AB = 8.4 cm. Find the measure of B, rounded to the nearest tenth of a degree.

10. A destroyer in the Atlantic Ocean is 2300 feet above the ocean’s floor when the captain notices an enemy submarine sitting on the ocean floor, 6200 feet away. At what angle of depression must the destroyer aim its missile to hit this enemy submarine?

5 11. ABC is a right triangle with right angle at vertex C. If the cosine of A is , find the 13 measure of B.

12. You are driving up a hill that is 2.8 miles long and 902.4 feet high. Find the angle of elevation of the hill.

13. For the figure pictured at right, find: a. the perimeter of the rectangle. 132” b. the area of the rectangle.

14. To find the height of the smokestack at New Trier, a student takes some measurements. The distance from where he is standing to the bottom of the smokestack is 110 feet, and, using a clinometer, he finds that the angle of elevation to the smokestack is 36.2°. How tall is the smokestack?

15. Find the angle  in the diagram.

350 feet

23° 1400 feet

19 16. A B C

13 F G

38° E D

For the figure given above, with rectangles ACDE and ABGF, find each of the following: a) the area of rectangle ABGF b) the length of DE c) the area of rectangle ACDE d) the area of trapezoid FGDE e) the area of trapezoid BCDG

17. A tent is in the shape of a cone with a diameter of 14 feet. The peak rises 8 feet above the center of the floor. How far from the edge must a 5-foot tall person stand in order to be able just barely touch the tent fabric with her head?

20 ANSWERS: Section I

1) a) opposite b) adjacent B B

adjacent opposite hypotenuse hypotenuse

2) a) i) adjacent = 3, opposite = 4, hypotenuse = 5 ii) sin() = 4/5, cos() = 3/5, tan() = 4/3 b) i) adjacent = 12, opposite = 5, hypotenuse = 13 ii) sin() = 5/13, cos() = 12/13, tan() = 5/12 c) i) adjacent = 12, opp = 6, hypotenuse ≈ 13.42 ii) sin() ≈ .4472, cos() ≈ .8944, tan() = .5 d) i) adjacent ≈ 43.87, opp = 10, hypotenuse = 45 ii) sin() = 2/9, cos() ≈ .9750, tan() ≈ .2279

3) sin() = a/c, cos() = b/c, tan() = a/b

4) a) ≈ 4.38 b) ≈ 12.29 c) ≈ 4.37 d) ≈ 38.46

5) a) ≈ 2.52 b) ≈ 2.96 c) ≈ 5.66

6) a) B 6b) 4/5 6c) ≈ .5547 6d) ≈ 1.4361

C A 7) a) 1/2 b) 1/2 c) 0.87 d) 0.87 e) 0.58 f) 1.73

8) a) 0.71 b) 0.71 c) 1 d) an isosceles triangle has two equal sides, so the opposite and adjacent are the same

9) ≈ 144.89 feet 10) ≈ 163.18 feet 11) ≈ 1348.58 feet

12) ≈ 103.92 feet 13) ≈ 20323 feet 14) ≈ 7159.06 feet

15) ≈ 5804.73 feet 16) right rear to curb ≈ 9.9 feet; left front to curve ≈ 4.24 feet

17) ≈ 104.43 feet 18) perim ≈ 89.91 19) ≈ 19446.9 kilometers

Section II

1) a) ≈ 30.96° b) ≈ 36.87° c) ≈ 55.15° d) 45°

2) x ≈ 7.71, y ≈ 18.75° 3) x ≈ 35.66°, y ≈ 71.32°

4) A = C ≈ 73.40°; ABC ≈ 33.20° 5) ≈ 84.56° 6) ≈ 10.75°

7) ≈ 41.30° 8) ≈ 38.66° 9) ≈ 30.9° 10) ≈ 21.78° 11) ≈ 22.62°

12) ≈ 3.5° 13) a) ≈ 350.84 units b) ≈ 6673.78 square units

14) ≈ 80.51 feet 15) ≈ 11°

16) a) 117 b) ≈ 20.68 c) ≈ 310.19 d) ≈ 101.04 e) ≈ 92.16

17) 4.375 feet from the edge of the tent