Trigonometry

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What You’ll Learn

• Determine the measure of an acute angle in a right triangle using the lengths of two sides.

• Determine the length of a side in a right triangle using the length of another side and the measure of an acute angle.

• Solve problems that involve more than one right triangle.

Why It’s Important

Trigonometric ratios are used by:

• surveyors, to determine the distance across a river or a very busy street

• pilots, to determine flight paths and measure crosswinds

• forestry technicians, to calculate the heights of trees

Key WordsDRAFT tangent ratio cosine ratio angle of inclination angle of elevation indirect measurement angle of depression sine ratio

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2.1 Skill Builder

Similar Triangles

Similar triangles have: • the measures of matching angles equal OR • the ratios of matching sides equal E 12 cm P S ° Q 30 5 cm 8 cm F 6 cm 4 cm A 110° T R U 2 cm 40° These triangles are not similar because the 40° ratios of matching sides are different. 110° 30° PQ 12 C B D 2.4 ST 5 These triangles are similar because QR 6 3 Compare the longest sides, matching angles are equal. TU 2 compare the shortest sides, A D 40° RP 8 then compare the third 2 pair of sides. B E 30° US 4 C F 110°

Check

1. Which triangles in each pair are similar?

B 1.0 cm Y a) M b) K 3 cm E F

3.0 cm 4 cm 5 cm 8 cm 3.5 cm 5.0 cm 10 cm 4.0 cm G

DRAFT W J 6 cm D C X 2.0 cm Compare the ratios of matching sides. Compare the ratios of matching sides. DB ______5.0 ______FG ______5 ______1.4 ______0.5 ______MJ ______3.5 ______XY ______10 CD ______2.0 ______EF _____ 3 ______2 ______0.5 ______KM ______1.0 ______WX _____6 BC ______4.0 ______GE _____ 4 ______1.3 ______0.5 ______JK ______3.0 ______YW _____8 The triangles ______are not similar. The triangles ______are similar.

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2.1 The Tangent Ratio

FOCUS Use the tangent ratio to find an angle measure.

The Tangent Ratio

A An acute angle is less than 90°. side adjacent to ∠A

C B side opposite ∠A length of side opposite A If A is an acute angle in a right triangle, then tan A length of side adjacent to A

Example 1 Finding the Tangent Ratio

Find the tangent ratio for G. 18 E F

Solution

Draw an arc at G. opposite 18 The side opposite G is EF. E F The side adjacent to G isDRAFT GE. 10 length of side opposite G adjacent tan G length of side adjacent to G G EF tan G Substitute: EF 18 and GE 10 GE 18 The side opposite the right angle tan G is always the hypotenuse. 10 tan G 1.8

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Check

1. a) Find tan P. b) Find tan Q. P The side opposite P is ______QR . The side opposite The side adjacent to P is ______RP . Q is ______RP . 8 Q length of side opposite P The side adjacent 10 ______tan P to Q is QR . R length of side ______adjacent to P ______length of side opposite Є Q QR tan Q tan P ______length of side adjacent to Є Q ______RP 10 RP tan P tan Q ______8 ______QR 8 tan P ______1.25 tan Q ______10 tan Q ______0.8

To find the measure of an angle, use the tan1 key on a scientific calculator.

Example 2 Using the Tangent Ratio to Find the Measure of an Angle

Find the measure of A to the nearest degree. 16 CB

Solution

The side opposite A is BC. The side adjacent to A is AB.DRAFT length of side opposite A tan A length of side adjacent to A If you are using a different BC tan A Substitute: BC 16 and AB 7 calculator, consult the user’s AB manual. 16 tan A 7

To find A using a TI-30XIIS calculator, enter: tan-1(16/7) 66.37062227 %@.3W4E<, A 66°

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Check

1. Find the measure of each indicated angle to the nearest degree.

a) F F

H The side opposite F is ______GH . The side adjacent to F is ______FG .

length of side ______opposite F tan F length of side ______adjacent to F G H tan F ______FG

13 tan F ______10 tan F ______1.3 Use a calculator.

F tan1______(1.3) F ______52°

b) E The side opposite E is ______CD . 5 The side adjacent to E is ______DE . CD length of side opposite ЄE tan E DRAFTlength of side adjacent to ЄE ______9 CD tan E ______DE 5 tan E E ______9

5 E tan1______aa 9bb

E ______29°

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Example 3 Using the Tangent Ratio to Find an Angle of Inclination

A guy wire is fastened to a cell-phone tower 8.5 m above the ground. The wire is anchored to the ground The angle the wire makes 14.0 m from the base of the tower. What angle, to the with the ground is called the angle of inclination. nearest degree, does the wire make with the ground? Solution A Draw a diagram. The angle the wire makes with the ground is B. Assume the tower is 8.5 m To find B, use the tangent ratio. perpendicular to the ground. B C 14.0 m length of side opposite B The side opposite B is CA. tan B length of side adjacent to B The side adjacent to B is BC. CA tan B Substitute: CA 8.5 and BC 14.0 BC 8.5 tan B Use a calculator. 14.0 B 31° The angle between the ground and the wire is about 31°.

Check

1. A ladder leans against a house. The top of the ladder is 2.4 m above the ground. F Its base is 0.9 m from the wall. What angle, to the nearest degree, does the ladder make with the ground?

Label the given triangle FGH. 2.4 m Label G where the ladder meets the ground. Label F where it meets the wall. We want to find the measureDRAFT of G. G H 0.9 m The side opposite G is ______.HF The side adjacent to G is ______GH .

HF opposite tan G tan G ______GH adjacent 2.4 tan G ______0.9 G ______69° The angle between the ground and the ladder is about ______69° .

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Practice

1. Label the hypotenuse, opposite, and adjacent sides of each right triangle in relation to the given angle. a) H b) P G M adjacent opposite opposite hypotenuse P hypotenuse N

F H adjacent

2. Find the tangent ratio for each indicated angle. Leave the ratio in fraction form. a) Z b) V 7 16 15 X 3 UW

The side opposite Y is ______ZX . The side opposite W is ______UV . The side adjacent to Y is ______XY . The side adjacent to W is ______VW . length of side ______opposite Y length of side opposite ЄW tan Y tan W length of side ______adjacent to Y ______length of side adjacent to ЄW

ZX UV tan Y tan W ______XY ______VW 7 tan Y tan W 16 ______3 DRAFT______15 3. Find the measure of A for each value of tan A. Give your answer to the nearest degree. 5 a) tan A 0.5 b) tan A 6 1 5 A tan (______0.5 ) Use a calculator. tan1 A ______aa 6bb A ______27° A ______40°

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4. Find the measure of ЄB to the nearest degree. 13 B C The side opposite ЄB is ______CD . The side adjacent to ЄB is ______BC .

12 length of side ______opposite Є B tan B ϭ length of side ______adjacent to Є B CD D tan B ϭ ______BC 12 tan B ϭ ______13 ЄB Џ ______43°

5. A telephone pole is supported by a wire, as shown. M What angle, to the nearest degree, does the wire make with the ground?

We want to find the measure of ЄN. 18 m Use the tangent ratio. length of side ______opposite Є N tan ______N ϭ Є N P length of side ______adjacent to N 8 m PM tan ______N ϭ ______NP 18 tan ______N ϭ ______8 Є______N Џ ______66°

The angle between the ground and the wire is about ______66° .

6. Victor is building a wheelchair ramp to an entranceway that is 3 m above the sidewalk. The ramp will cover a horizontal distance of 50 m. What angle, to the nearest degree, will the ramp make with the ground? R 3 m DRAFT Q TEACHER NOTE S 50 m Next Steps: Have We want to find the measure of ЄQ. students complete Use the tangent ratio. questions 6, 8, 10, 14, and 15 on pages 75 length of side opposite ЄQ and 76 of the Student ؍ tan Q length of side adjacent to ЄQ Text. For students RS ,experiencing success ؍ tan Q SQ introduce Example 4, 3 on page 74 of the ؍ tan Q 50 Student Text, and assign Practice Є Џ Q 3° questions 13, 19, The angle between the ground and the ramp is about ______3° . and 20.

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2.2 Skill Builder

Solving Equations

Inverse operations “undo” each other’s results. Multiplication and division are inverse operations. We can use inverse operations to solve some equations.

a a To solve 4: 4 Undo the division. 5 5 Multiply each side by 5. a 5 5 4 5 a 20

36 36 Undo the division. To solve 9: b b 9 b b Multiply each side by b. 36 9b Recall: b 9 9b 36 9b Undo the multiplication. 9 9 Divide each side by 9. 4 b

Check

1. Solve each equation. a c a) 5 b) 9 7 8 Multiply each side by ______7 . Multiply each side by ______8 . a c ______7 ______7 5 8 9 8 7 8 a ______35 DRAFT72 c 156 15 c) 13 d) 6 f b 15 Multiply each side by ______f . b b b 6 156 15 6b ______f 13 ______f f 15 6b ______13f ______156 6 6 2.5 b Divide each side by ______13 . 13f 156 ______13 ______13 ______f ______12

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2.2 Using the Tangent Ratio to Calculate Lengths

FOCUS Use the tangent ratio to calculate lengths.

When we know the measure of an acute angle and the length of a leg of a right triangle, we can use the tangent ratio to find the length of the other leg.

Example 1 Finding the Length of an Opposite Side

Find the length of BC to the nearest tenth of a centimetre.

23.0 cm CA 28°

Solution

We know the measure of ЄA. adjacent to ≠A 23.0 cm CA BC is the side opposite ЄA. 28° CA is the side adjacent to ЄA. opposite ≠A Use the tangent ratio to write an equation. length of side opposite Є A B tan A length of side adjacent to Є A BC tan A Substitute: ЄA 28° and CA 23 CA BC tan 28° 23 Solve the equation for BC. Multiply each side by 23. BC 23 tan 28° 23 DRAFT 23 23 tan 28° BC Use a calculator. Enter: / 0 V @ / 5 E <

BC 12.2293… 23*tan(28) 12.22931693 BC is about 12.2 cm long.

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Check

1. Find the length of each indicated side to the nearest tenth of a centimetre.

a) Side ED The given angle is ЄF. F ______DE is the side opposite ЄF. Є 59° ______EF is the side adjacent to F. side opposite Є F 8 cm ______tan F side ______adjacent to Є F D E DE tan F ______EF DE tan ______59° ______8

Solve the equation for DE. Multiply each side by ______8 . DE ______8 tan ______59° ______8 ______8

______8 tan ______59° ______DE DE ______13.3142…

DE is about ______13.3 cm long.

b) Side HJ The given angle is ______ЄK . J ______HJ is the side opposite ______ЄK . ______JK is the side adjacent to ______ЄK . 13 cm side opposite Є K H 34° tan ______K ______side adjacent to Є K K HJ tan ______K ______JK HJ DRAFTtan ______34° ______13 HJ 13 tan 34° 13 13 13 tan 34° HJ HJ 8.7686… HJ is about ______8.8 cm long.

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Example 2 Finding the Length of an Adjacent Side

Find the length of PQ to the nearest tenth of a centimetre. Q 5.0 cm

R 35° P

Solution

Use the tangent ratio to write an equation. Є side opposite Є P We know P 35°. tan P QR is opposite ЄP. side adjacent to ЄP PQ is adjacent ЄP. QR tan P Substitute: QR 5 and ЄP 35° PQ 5 tan 35° PQ Solve the equation for PQ. Multiply each side by PQ. 5 PQ tan 35° PQ PQ PQ tan 35° 5 Divide each side by tan 35°. PQ tan 35 5 tan 35 tan 35 5 PQ Use a calculator. tan 35 PQ 7.1407… So, PQ is about 7.1 cm long.DRAFT

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Check

1. Find the length of TU to the nearest tenth of a centimetre.

4.0 cm

39° U T

The given angle is ЄT. ______US is the side opposite ЄT. ______TU is the side adjacent to ЄT. side ______opposite Є T tan T side ______adjacent to Є T US Є tan T ______TU Substitute: ______T 39° and ______US 4 4 tan ______39° ______TU Multiply each side by ______TU . 4 ______TU tan ______39° ______TU ______TU

______TU tan ______39° ______4 Divide each side by ______tan 39° .

_____ T U tan _____ 39° 4 ______tan 39° ______tan 39° 4 TU Use a calculator. ______tan 39Њ

TU ______4.9395… TU is about ______4.9 cm long.DRAFT

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Example 3 Using the Tangent Ratio to Solve a Problem

A wire supports a flagpole. The angle between the wire and the level ground is 73°. The wire is anchored to the ground 10 m from the base of the pole. How high up the pole does the wire reach? Give the answer to the nearest tenth of a metre.

Solution

Sketch and label a diagram. Assume the flagpole meets the ground at a right angle. B

The given angle is ЄA. We want to find the length of BC. side opposite ЄA tan A side adjacent to ЄA

BC is opposite ЄA. CA is adjacent to ЄA. 73° A C 10.0 m BC tan A Substitute: ЄA 73° and CA 10 CA Solve the equation for BC. BC tan 73° Multiply each side by 10. 10 10 tan 73° BC Use a calculator.

BC 32.7085…

The wire reaches the flagpole at a height of about 32.7 m.

Check

1. A ladder leans on a wall, as shown. How far up the wall does the ladder reach? Give yourDRAFT answer to the nearest tenth of a metre.

The given angle is ЄF. D We want to find the length of ______DE .

67° E F 1.5 m

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side opposite Є F ______tan F side ______adjacent to Є F DE tan F ______EF Substitute: ______ЄF 67° and ______EF 1.5

DE tan ______67° ______1.5 Multiply each side by ______1.5 . ______1.5 tan ______67° ______DE DE ______3.5337…

The ladder reaches the wall at a height of about ______3.5 m .

Practice

1. Find the length of the side opposite the given angle to the nearest tenth of a centimetre. a) G b) Z

X 57°

5 cm 31° F H 6 cm Y

The given angle is ЄF. The given angle is Є______X . The side opposite ЄF is ______GH . The side opposite Є______X is ______YZ . The side adjacent to ЄF is ______HF . The side adjacent to Є______X is ______XY .

side ______opposite Є F side ______opposite Є X tan F tan ______X side ______adjacent to Є F side ______adjacent to Є X

GH YZ tan F tan X ______HF XY DRAFTYZ GH tan 57° 5 tan ______31° ______6 YZ 5 tan 57° 5 GH 5 6 5 tan 57° YZ ______6 tan ______31° ______6 YZ 7.6993… ______6 tan ______31° ______GH ______YZ is about ______7.7 cm long. GH ______3.6051… GH is about ______3.6 cm long.

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2. Find the length of CD to the nearest tenth of a centimetre. D Є 16 cm The given angle is C. Є DE E The side opposite C is ______. The side adjacent to ЄC is ______CD . 44° side ______opposite Є C C tan C side ______adjacent to Є C

DE tan C ______CD 16 tan ______44° ______CD Multiply each side by ______CD . 16 CD tan 44° CD CD CD tan 44° 16 Divide each side by ______tan 44° . CD tan 44° 16 tan 44° tan 44Њ 16 CD tan 44Њ CD 16.5684… CD is about ______16.6 cm long.

3. Find the length of the indicated side to the nearest tenth of a centimetre.

a) Side PQ b) Side UV P V

Q 32 cm U 60°

18 cm T 36° DRAFT opposite tan U adjacent R ______opposite 32 tan R tan 60° UV adjacent ______32 PQ UV tan 60° UV UV tan ______36° ______18 UV tan 60° 32 PQ 18 tan 36° 18 Њ 18 UV tan 60 32 tan 60Њ tan 60Њ 18 tan 36° PQ UV 18.4752… PQ 13.0777… UV is about ______18.5 cm long. PQ is about ______13.1 cm long. 16 MMS10_L2.qxp 1/21/10 10:36 AM Page 17

4. This diagram shows an awning over the window H 32° of a house. Find the height of the awning, GH, to the nearest tenth of a metre. opposite tan H ______adjacent J 1.6 G 1.6 m tan 32° GH 1.6 window GH tan 32° GH GH GH tan 32° 1.6 GH tan 32Њ 1.6 tan 32Њ tan 32Њ GH 2.5605… The height of the awning is about ______2.6 m .

5. A rope supports a tent. The angle between the rope and the level ground is 59°. The rope is attached to the ground 1.2 m from the base of the tent. At what height above the ground is the rope attached to the tent? Give your answer to the nearest tenth of a metre.

59° A C 1.2 m

We want to find the length of BC. opposite tan A adjacent BC tan A AC BC tan 59° DRAFT 1.2 TEACHER NOTE BC 1.2 tan 59° 1.2 Next Steps: Have 1.2 students complete 1.2 tan 59° BC questions 6, 8, 10, 11, BC 1.9971… and 13 on pages 82 and 83 of the Student The rope is attached to the tent at a height of about ______2.0 m . Text.

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2.3 Math Lab: Measuring an Inaccessible Height

FOCUS Determine a height that cannot be measured directly.

When we find a length or an angle without using a measuring instrument, we are using indirect measurement.

Try This Work with a partner. Follow the instructions in Part A on Student Text page 85 to make a clinometer. The materials you need are listed on Student Text page 84. Record all your measurements on the diagram below. Choose a tall object; for example, a tree or a flagpole. Object: ______Mark a point on the ground. Measure the distance to the base of the object. One person stands at the point. He holds the clinometer, then looks at the top of the object through the straw. The other person records the angle shown by the thread on the protractor. Then that person measures the height of the eyes above the ground of the person holding the clinometer. Subtract the clinometer angle from 90°. This is the angle of inclination of the straw.

Clinometer angle: DRAFT Height of object: Angle of inclination:

A C

Height of eyes:

Distance to object:

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Use the tangent ratio to calculate the length of BC: BC tan A AC BC tan ______ ______

BC ______ tan ______BC Џ ______

Height of object length of BC height of eyes above the ground Height of object ______ ______Height of object ______

Change places with your partner. Repeat the activity. Does the height of your eyes affect the measurements? Explain. Sample response: Yes,______the height of your eyes affects the angle of inclination of the object. ______The______taller you are, the less you have to look up to the object. ______

Does the height of your eyes affect the final result? Explain. Sample response: No,______a taller person will calculate a greater height above the horizontal but will ______have a ______lesser eye height. But, when the two heights are added, both results should be______the same.______

Practice

1. Which angle of inclination does each clinometer measure?

180 a) 180DRAFTb)

160 160

140

120

120 40

100

0 100

20 80

40 60

Angle of inclination Angle of inclination 90° angle on clinometer 90° ______80° 90° ______55° ______10° ______35° 19 MMS10_L3.qxp 1/21/10 5:08 PM Page 20

2. Use the information in the diagram to find 180 the height of the flagpole to the nearest tenth of a metre. 160 60°

F 140

Angle on clinometer: _____60° 0

120

Angle of inclination: 90° Ϫ _____60° ϭ _____30° 20

100

80 So, ЄE ϭ ______30° 60 opposite ϭ tan E E adjacent G

FG 1.3 m tan ______30° ϭ ______EG 4.0 m FG tan ______30° ϭ ______4 FG 4 ؍ tan 30° 4 aa 4 bb FG ؍ tan 30° 4 FG ϭ ______2.3094… So, height of flagpole ϭ ______FG ϩ height of eyes above ground ϭ ______2.3094… ϩ ______1.3 ϭ ______3.6094… The height of the flagpole is about ______3.6 m .

3. Use the information in the diagram to find 180 75°

the height of the tree to the nearest tenth 160

of a metre. 140 120

Є Ϫ ϭ 20 40

P is: 90° ______75° ______15° 100

60 80 opposite tan P ϭ Q ______adjacentDRAFT QR P R tan ______15° ϭ ______PR 1.7 m QR m 10 ؍ tan 15° 10 This diagram is not drawn to scale. QR ؍ tan 15° 10 QR ϭ ______2.6794… TEACHER NOTE So, height of tree ϭ 2.6794… ؉ 1.7 ______Next Steps: Have ϭ ______4.3794… students complete question 1 on page 86 The height of the tree is about ______4.4 m . of the Student Text.

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CHECKPOINT 1 Can you …

• use the tangent ratio to find an angle measure? • use the tangent ratio to calculate a length? • use the tangent ratio to solve a problem?

2.1 1. Find the tangent ratio for each indicated angle. Leave the ratio in fraction form.

11 E a) B C b) 22 7 15 D

A F Є The side opposite A is ______.BC side opposite Є D tan D The side adjacent to ЄA is ______AB . ______side adjacent to Є D side ______opposite Є A EF tan A tan D DE side ______adjacent to Є A ______BC tan D 15 tan A ______AB ______22 11 tan A ______7

2. Find the measure of each indicated angle to the nearest degree.

a) G b) K 8 17 7 J F H 22 The side opposite ЄH is ______FG . M The side adjacent to ЄDRAFTH is ______GH . side opposite ЄK side opposite Є H ______tan H tan _____K ______side adjacent to Є K side ______adjacent to Є H MJ FG tan _____K ______JK tan H ______GH 22 8 tan H tan _____K ______7 ______17 ЄK Џ ______72° ЄH Џ ______25°

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2.2 3. Find the length of each indicated side to the nearest tenth of a centimetre.

a) Side ST b) Side PQ T 24 cm P R 21 cm

38° U 68° Q side opposite Є Q S tan ______Q ______side adjacent to Є Q side ______opposite Є U RP tan U tan Q PQ side ______adjacent to Є U 24 ST tan 68° PQ tan U TU ______24 PQ tan 68° PQ ST PQ tan ______38° ______21 PQ tan 68° 24 ST 21 tan 38° 21 21 Њ PQ tan 68 24 21 tan 38° ST tan 68Њ tan 68° ST ______16.4069… PQ 9.6966… ST is about ______16.4 cm long. PQ is about ______9.7 cm long.

4. Margy is building a support brace to reach the top of a wall, as shown. C How far from the wall should the brace be anchored to the ground? Give your answer to the nearest tenth of a metre.

We want to find the length of AB. 3.5 m The side opposite ЄA is BC. The side adjacent to ЄA is AB. 70° side opposite Є A BA tan A side adjacent to Є A BC tan A AB DRAFT 3.5 tan 70° AB 3.5 AB tan 70° AB AB AB tan 70° 3.5 AB tan 70Њ 3.5 tan 70Њ tan 70Њ TEACHER NOTE 3.5 Next Steps: Have AB students complete tan 70Њ questions 3 and 5 AB 1.2738… on page 88 of the The brace should be anchored to the ground ______about 1.3 m from the wall. Student Text.

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2.4 Skill Builder

Sum of the Angles in a Triangle

In any triangle, the sum of the In any right triangle, the sum of the angle measures is 180°. measures of the acute angles is 90°. So, to find an unknown angle measure: So, to find the measure of an acute angle: • start with 180° • start with 90° • subtract the known measures • subtract the known acute angle A D 40° E 71°

64° B

C F ЄC 180° 71° 64° ЄE 90° 40° ЄC 45° ЄE = 50°

Check

1. Find the measure of the third angle.

a) H b) M

O 59° G 46° 30° 76° J N

ЄH 180° ______46° ______30° ЄM ______180° ؊ 76° ؊ 59° ЄH ______104° DRAFTЄM ______45° 2. Find the measure of the third angle.

a) T b) Q 56°

49° R

S U P

ЄU 90° ______56° ЄP ______90° ؊ 49° ЄU ______34° ЄP ______41°

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2.4 The Sine and Cosine Ratios

FOCUS Use the sine and cosine ratios to determine angle measures.

C The Sine and Cosine Ratios In a right triangle, if ЄA is an acute angle, then length of side opposite Є A sin A side opposite ∠A hypotenuse length of hypotenuse length of side adjacent to Є A cos A length of hypotenuse B A side adjacent to ∠A

Example 1 Finding the Sine and Cosine of an Angle

Find sin B and cos B to the nearest hundredth. A 26 cm

B The nearest hundredth 24 cm means two decimal places. 10 cm

C Solution

AC is the side opposite ЄB. A hypotenuse 26 cm AB is the hypotenuse. B length of side opposite Є B 24 cm sin B side opposite ∠B length of hypotenuse 10 cm side adjacent to ∠B AC sin B Substitute: AC 24 and AB 26 C AB 24 sin B 26 sin B 0.9230… DRAFT sin B Џ 0.92 BC is the side adjacent to ЄB. AB is the hypotenuse. length of side adjacent to Є B cos B length of hypotenuse BC cos B Substitute: BC 10 and AB 26 AB 10 cos B 26 cos B 0.3846… cos B Џ 0.38

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Check

1. Find sin D and cos D to the nearest hundredth. D

side opposite Є D side adjacent to Є D 7.5 cm sin D cos D hypotenuse hypotenuse F 8.5 cm EF FD 4.0 cm sin D ______DE cos D ______DE

4.0 7.5 E sin D ______8.5 cos D ______8.5 sin D ______0.4705… cos D ______0.8823…

sin D Џ ______0.47 cos D Џ ______0.88

To find the measure of an angle, use the sin1 or cos1 key on a scientific calculator.

Example 2 Using the Sine or Cosine Ratio to Find the Measure of an Angle

Find the measures of ЄB and ЄD C

to the nearest degree. 4.6 cm D B 8.7 cm Solution

Find the measure of ЄB first. BC is adjacent to ЄB. BD is the hypotenuse. So, use the cosine ratio to write an equation. We could use the sine ratio to find the measure of ЄD side adjacent to ЄB first. cos B hypotenuse BC cos B DRAFTSubstitute: BC 4.6 and BD 8.7 BD 4.6 cos B 8.7 To find ЄB using a TI-30XIIS calculator, enter:

% ? 1 8 3 W 5 8 4 E < cos-1(4.6/8.7) 58.07993367 ЄB Џ 58°

Since the sum of the acute angles in a right triangle is 90°, ЄD 90° ЄB Since the measure of ЄB Є Џ is an estimate, so is the So, D 90° 58° measure of ЄD. ЄD Џ 32°

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Check

1. Find the measure of each acute angle to the nearest degree.

5 cm a) U V Find the measure of ЄU first. UV is the ______side adjacent to ЄU. UW is the ______hypotenuse . 9 cm So, use the cosine ratio to write an equation.

side adjacent to U cos U hypotenuse W ______UV cos U ______UW 5 cos U ______9 ЄU Џ ______56° So, ЄW 90° ЄU ЄW Џ 90° ______56° ЄW Џ ______34°

b) T Find the measure of ЄU first. 10.8 cm ST is the ______side opposite U . S SU is the ______hypotenuse . So, use the ______sine ratio to write an equation. 18.0 cm side opposite U U ______sin U ______hypotenuse

ST ______sin U ______US

10.8 DRAFT______sin U ______18.0 ЄU Џ ______37° So, ЄS 90° ______U ЄS Џ 90° ______37° ЄS Џ ______53°

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Example 3 Using Sine or Cosine to Solve a Problem

A storm caused a 15.3-m hydro pole to lean over. The top of the pole is now 12.0 m above the ground. What angle does the pole make with the ground? Give the answer to the nearest degree.

Solution

Draw a diagram. AC represents the pole. C The pole meets the ground at A.

BC is the side opposite ЄA. AC is the hypotenuse. 15.3 m 12.0 m So, use the sine ratio to find ЄA.

side opposite Є A A B sin A hypotenuse BC sin A Substitute: BC 12.0 and AC 15.3 Assume the ground is AC horizontal. 12.0 sin A Use a calculator. 15.3 ЄA Џ 52° So, the hydro pole makes an angle of about 52° with the ground.

Check

1. A ladder leans on a wall as shown. What angle does the ladder make with the ground? E Give your answer to the nearest degree. We want to find the measure of ЄD. DF is ______adjacent to D . DE is ______the hypotenuseDRAFT. 8 m So, use the ______cosine ratio to find ЄD.

D 2 m F

Assume the ground is horizontal.

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side adjacent to D ______cos D ______hypotenuse

DF ؍ ؍ ______cos D ______DE Substitute: ______DF 2 and ______DE 8

2 ______cos D ______8 ЄD Џ ______76° So, the ladder makes an angle of about ______76° with the ground.

2. The string of a kite is 160 m long. The string is anchored to the ground. The kite is 148 m high. What angle does the string make U with the ground? Give your answer to the nearest degree. 148 m 160 m We want to find the measure of ЄV. TU is the side opposite V . ______T V UV is the ______hypotenuse . So, use the ______sine ratio to write an equation. side opposite V ؍ sin V hypotenuse TU ؍ sin V UV 148 ؍ sin V 160 V 68° The angle the string makes with the ground is about ______68° .

Practice DRAFT 1. Fill in the blanks. a) C b) G 5.0 4 3 B 4.8 D 1.4 B 5 F The side opposite ЄB is ______CD . The side opposite ЄB is ______FG . The side adjacent to ЄB is ______BC . The side adjacent to ЄB is ______BF . The hypotenuse is ______BD . The hypotenuse is ______GB .

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2. For each triangle in question 1, find sin B and cos B as decimals.

side ______opposite Є B side ______adjacent to Є B a) sin B cos B hypotenuse hypotenuse CD BC sin B ______BD cos B ______BD 4 3 sin B ______5 cos B ______5 sin B ______0.8 cos B ______0.6

side opposite B side adjacent to B b) sin B ______hypotenuse cos B ______hypotenuse

FG BF ؍ cos B ؍ sin B GB GB 4.8 1.4 ؍ cos B ؍ sin B 5.0 5.0

sin B ______0.96 cos B ______0.28

3. Find the measure of each indicated angle to the nearest degree.

a) A 6.9 cm b) X B 9.9 cm 5.5 cm Y

C W 11.7 cm

AC is the ______side opposite B . XY is the ______.side adjacent to Y AB is the ______hypotenuse . WY is the ______hypotenuse . So, use the ______sine ratio. So, use the ______cosine ratio.

side opposite B side adjacent to Y ؍ DRAFTcos Y ؍ sin B hypotenuse hypotenuse AC X Y ؍ cos Y ؍ sin B AB WY 5.5 9.9 ؍ cos Y ؍ sin B 6.9 11.7 ЄB Џ ______53° ЄY Џ ______32°

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4. A firefighter rests a 15.6-m ladder against a building, as shown. G What angle does the ladder make with the ground? Give your answer to the nearest degree. 15.6 m We want to find the measure of ЄH. FH is the ______side adjacent to ЄH . GH is the ______hypotenuse . H So, use the ______cosine ratio. F 8.5 m

side adjacent to ЄH ______cos H ϭ ______hypotenuse

FH ______cos H ϭ ______GH

8.5 ______cos H ϭ ______15.6 ЄH Џ ______57° The angle the ladder makes with the ground is about ______57° .

5. A loading ramp is 4.5 m long. The top of the ramp has height 1.6 m. What angle does the ramp make with the ground? Give your answer to the nearest degree.

N 4.5 m 1.6 m

M P

We want to find the measure of ЄM. NP is the side opposite ЄM. MN is the hypotenuse. So, use the sine ratio. DRAFT side opposite ЄM ؍ sin M hypotenuse NP ؍ sin M MN TEACHER NOTE Next Steps: Have 1.6 students complete ؍ sin M 4.5 questions 7, 8, 10, 11, ЄM Џ 21° 13, and 14 on pages 95 and 96 of the Student The angle the ramp makes with the ground is about ______21° . Text.

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2.5 Using the Sine and Cosine Ratios to Calculate Lengths

FOCUS Use the sine and cosine ratios to determine lengths.

To use the sine or cosine ratio to find the length of a leg, we need to know: • the measure of an acute angle, and • the length of the hypotenuse

Example 1 Using the Sine or Cosine Ratio to Find the Length of a Leg

Find the length of RS to the nearest tenth of a metre.

R S 28°

9.6 m

Solution

The measure of ЄS is known. adjacent to ≠S R ° S RS is the side adjacent to ЄS. 28 QS is the hypotenuse. 9.6 m So, use the cosine ratio. hypotenuse Q side adjacent to Є S cos S hypotenuse RS cos S Substitute: ЄS 28° and QS 9.6 QS RS cos 28° Multiply both sides by 9.6. The cosine ratio compares 9.6 the adjacent side to the RS 9.6 cos 28° 9.6 DRAFT hypotenuse. 9.6 9.6 cos 28° RS Use a calculator. RS 8.4762… RS is about 8.5 m long.

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Check

1. Find the length of each indicated side to the nearest tenth of a centimetre.

a) AC The measure of ЄB is known. B AC is the ______side opposite ЄB . 40° BC is the ______hypotenuse . So, use the ______sine ratio. 15.6 cm side ______opposite ЄB ______sin B hypotenuse AC A C ______sin B BC AC sin 40° 15.6 AC 15.6 sin 40° 15.6 15.6 15.6 sin 40° AC AC ______10.0274… AC is about ______10.0 cm long.

b) DE The measure of Є______D is known. E DE is the ______side adjacent to ЄD . DF is the ______hypotenuse . F So, use the ______cosine ratio.

55° 19.5 cm side adjacent to ЄD D cos D hypotenuse DE cos D DF DE cos 55° 19.5 DRAFT19.5 cos 55° DE DE 11.1847… DE is about ______11.2 cm long.

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To use the sine or cosine ratio to find the length of the hypotenuse, we need to know: • the measure of an acute angle, and • the length of one leg

Example 2 Using the Sine or Cosine Ratio to Find the Length of the Hypotenuse

Find the length of the hypotenuse to the nearest tenth of a centimetre.

9.5 cm The hypotenuse is the side opposite the right angle.

52° MP

Solution

The measure of ЄM is known. N NP is the side opposite ЄM. MN is the hypotenuse. opposite ≠M hypotenuse So, use the sine ratio to write an equation. 9.5 cm side opposite Є M sin M 52° hypotenuse M P NP sin M Substitute: ЄM 52° and NP 9.5 MN 9.5 sin 52° Multiply both sides by MN. MN The sine ratio compares MN sin 52° 9.5 Divide both sides by sin 52°. the opposite side to the hypotenuse. MN sin 52 9.5 sin 52 sin 52 DRAFT 9.5 MN Use a calculator. sin 52 MN 12.0556… MN is about 12.1 cm long.

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Check

1. Find the length of each hypotenuse to the nearest tenth of a centimetre.

a) K side ______opposite ЄJ sin J ______hypotenuse 17.4 cm KM M sin J ______JK 39° 17.4 J sin _____39° ______JK The measure of ЄJ is known. _____JK sin _____39° ______17.4 The side opposite ЄJ is: ______KM The hypotenuse is: ______JK 17.4 JK Њ Use the sine ratio. ______sin 39 JK ______27.6488… JK is about ______27.6 cm long.

b) S side adjacent to Є S ؍ cos S 29° R hypotenuse

11.9 cm QS ؍ cos S RS Q 11.9 ؍ cos 29° The measure of Є______S is known. RS QS is the side ______adjacent to ЄS . 11.9 ؍ RS cos 29° RS is the ______hypotenuse . 11.9 ؍ So, use the ______cosine ratio. RS cos 29Њ RS ______13.6059… DRAFTRS is about ______13.6 cm long.

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Example 3 Using Sine or Cosine to Solve a Problem

A surveyor makes the measurements shown in the diagram to find the distance between two observation towers on opposite sides of a river. How far apart are the towers? Give the answer to the nearest metre.

River

Tower 1 63 m 73°

Tower 2

Solution

The distance between the towers is the hypotenuse, AC. A B Tower 1 63 m 73° C Tower 2

The measure of ЄC is known. BC is the side adjacent to ЄC. AC is the hypotenuse. So, use the cosine ratio. side adjacent to ЄC cos C hypotenuse BC cos C Substitute: ЄC 73° and BC 63 AC 63 cos 73° Multiply both sides by AC. AC AC cos 73° 63 DRAFTDivide both sides by cos 73°. 63 AC Use a calculator. cos 73 AC 215.4791… The distance between the towers is about 215 m.

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Check

1. Sam and Sofia are building a wooden ramp for skateboarding. The height of the ramp is 0.75 m. The ramp makes an angle of 8° with the ground. What length of plywood do Sam and Sofia need for the top of the ramp? Give your answer to the nearest tenth of a metre.

0.75 m D 8° F

We want to find the length of DE. The measure of ЄD is known. The side opposite ЄD is: _____EF The hypotenuse is: _____DE So, use the sine ratio. EF Є sin D ______DE Substitute: ______D 8° and EF 0.75 0.75 sin _____8° ______DE Multiply both sides by ______DE .

_____DE sin _____8° _____0.75 Divide both sides by ______sin 8° .

0.75 DE sin 8Њ

DE ______5.3889… Sam and Sofia need about ______5.4 m of plywood.

Practice DRAFT 1. Which ratio would you use to find each length? a) XY b) ST X T

53° U ° ZY29 S 15.9 cm 17.6 cm The measure of Є______Y is known. The measure of Є______U is known. YZ is the side ______adjacent to ЄY . ST is the side ______opposite ЄU . XY is the ______hypotenuse . SU is the ______hypotenuse . So, use the ______cosine ratio. So, use the ______sine ratio. 36 MMS10_L5.qxp 1/21/10 10:40 AM Page 37

2. Find the length of each indicated side to the nearest tenth of a centimetre. a) VW

V The measure of Є______U is known. The side opposite Є______U is ______VW . The hypotenuse is ______UW . 48° U So, use the ______sine ratio. 15.0 cm W side ______opposite Є U sin ______U hypotenuse

VW sin ______U ______U W

VW sin _____48° ______15.0 ______15.0 sin ______48° ______VW VW ______11.1471…

VW is about ______11.1 cm long.

b) QR

R The measure of Є______S is known. QR is the ______side opposite ЄS . 10.5 cm RS is the ______hypotenuse . So, use the ______sine ratio. Q 36° S side opposite Є S sin S hypotenuse QR sin S RS

QR sin 36° DRAFT10.5 10.5 sin 36° QR QR 6.1717… QR is about ______6.2 cm long.

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3. Find the length of side PM to the nearest tenth of a metre.

P M The measure of Є______M is known. 19° PM is the side ______adjacent to ЄM . 16.0 m N MN is the ______hypotenuse . So, use the ______cosine ratio.

cos M P M MN PM cos 19° 16.0 16.0 cos 19° PM PM 15.1282… PM is about ______15.1 m long.

4. Find the length of each hypotenuse to the nearest tenth of a centimetre. a) E b) X 29° W 33°

21.6 cm Y

The measure of Є______W is known. WY is the side ______adjacent to ЄW .

GF WX is the ______hypotenuse . 8.4 cm So, use the ______cosine ratio. The measure of Є E is known. ______WY Є cos W The side opposite ______E is: ______FG WX The hypotenuse is: ______EF 21.6 So, use the sine ratio. cos 33° WX FG WX cos 33° 21.6 sin ______E ______EF 21.6 8.4 WX cos 33Њ sin ______29° ______EFDRAFT WX ______25.7550… ______EF sin ______29° ______8.4 WX is about ______25.8 cm long. 8.4 EF ______sin 29Њ EF ______17.3263…

EF is about ______17.3 cm long.

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5. A straight slide in a playground makes an angle of 28° with R the ground. The slide covers a horizontal distance of 4.5 m. How long is the slide? Give your answer to the nearest tenth of a metre. 28° Q 4.5 m S The measure of ЄQ is known. The side adjacent to ЄQ is: ______QS The hypotenuse is: ______QR So, use the ______cosine ratio.

QS ؍ cos Q QR

4.5 ؍ cos 28° QR

4.5 ؍ QR cos 28° 4.5 ؍ QR cos 28Њ

QR ϭ ______5.0965…

The slide is about ______5.1 m long.

6. A 15-m support cable joins the top of a telephone pole C to a point on the ground. The cable makes an angle of 32° with the ground. Find the height of the pole to the nearest tenth 15.0 m of a metre. ° E 32 Use the sine ratio. D

CE ؍ sin D CD

CE ؍ sin 32° 15.0 TEACHER NOTE CE DRAFT Next Steps: Have ؍ sin 32° 15.0 students complete questions 6–11 on …7.9487 ؍ CE pages 101 and 102 of The height of the pole is about ______7.9 m . the Student Text.

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CHECKPOINT 2 Can you …

• use the sine or cosine ratio to find an angle measure? • use the sine or cosine ratio to calculate a length? • use the sine or cosine ratio to solve a problem?

2.4 1. Find sin A and cos A to the nearest hundredth. B

13.0 16.0

A 20.6 C

side opposite ЄA side adjacent to Є A sin A ______hypotenuse cos A ______hypotenuse

BC AB sin A ______AC cos A ______AC

16.0 13.0 sin A ______20.6 cos A ______20.6

sin A ______0.7766… cos A ______0.6310… sin A Џ ______0.78 cos A Џ ______0.63

2. Find the measure of each indicated angle to the nearest degree.

17 cm 13.4 cm a) FD b) G H

5.9 cm 23 cm DRAFTJ E GJ is the ______side opposite ЄH . FD is the ______.side adjacent to ЄD GH is the ______hypotenuse . DE is the ______hypotenuse . So, use the ______sine ratio. So, use the ______cosine ratio. side opposite ЄH side adjacent to ЄD ______sin H ______hypotenuse ______cos D ______hypotenuse GJ FD ______sin H ______GH cos D DE ______5.9 17 ______sin H ______13.4 ______cos D ______23 ЄH Џ ______26° ЄD Џ ______42° 40 MMS10_L5.qxp 1/21/10 10:40 AM Page 41

2.5 3. Find the length of each indicated side to the nearest tenth of a centimetre. a) PR b) ST Q T

° 40° S 78

3.7 cm 22.8 cm U

The measure of Є______S is known. P R ST is the side ______adjacent to ЄS . The measure of Є______Q is known. SU is the ______hypotenuse . The side opposite Є______Q is ______PR . So, use the ______cosine ratio. side adjacent to ЄS The hypotenuse is ______PQ . cos S hypotenuse So, use the ______sine ratio. ST side opposite ЄQ cos S sin Q SU hypotenuse ST cos 78° PR 22.8 sin Q PQ 22.8 cos 78° ST P R sin 40° ST 4.7403… 3.7 ST is about ______4.7 cm long. 3.7 sin 40° PR PR 2.3783… PR is about ______2.4 cm long.

4. Find the length of the hypotenuse X 4.4 cm to the nearest tenth of a centimetre. Y The measure of ЄW is known. The side opposite ЄW is: ______XY 28° The hypotenuse is: ______WY W Use the sine ratio. DRAFTXY sin W ______W Y 4.4 sin ______28° ______WY Multiply both sides by ______WY . ______WY sin ______28° ______4.4 Divide both sides by ______.sin 28° 4.4 TEACHER NOTE WY sin 28Њ Next Steps: Have students complete WY 9.3722… questions 2 and 5 on page 104 of the WY is about ______9.4 cm long. Student Text.

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2.6 Applying the Trigonometric Ratios

FOCUS Use trigonometric ratios to solve a right triangle.

When we solve a triangle, we find the measures of all the angles and the lengths of all the sides. To do this we use any of the sine, cosine, and tangent ratios. hypotenuse A B We can use the acronym side adjacent to ∠A side opposite ∠A SOH-CAH-TOA to help us remember these ratios. C

opposite adjacent opposite sin A cos A tan A hypotenuse hypotenuse adjacent

Example 1 Finding the Measures of All Angles

Find all unknown angle measures to the nearest degree. E

5 cm 9 cm

D F

Solution

Find the measure of ЄD. EF is the side opposite ЄD. We could have used the tangent ratio to find ЄF. DE is the side adjacent to ЄD. So, use the tangent ratio. opposite tan D adjacent DRAFT EF tan D Substitute: EF 9 and DE 5 DE 9 tan D 5 ЄD Џ 61° The acute angles in a right triangle have a sum of 90°. So, ЄF 90° ЄD ЄF Џ 90° 61° ЄF Џ 29°

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Check

1. Find all unknown angle measures to the nearest degree.

a) H opposite ______tan G ______adjacent 10.5 cm 6.4 cm HJ J ______tan G ______GH G 6.4 Є Find the measure of G. ______tan G ______10.5 HJ is the side ______opposite ЄG . ЄG Џ ______31° GH is the side ______adjacent to ЄG . The acute angles have a sum of 90°. So, use the ______tangent ratio. So, ЄJ 90° ______ЄG ЄJ Џ 90° ______31° We could use the tangent ЄJ Џ ______59° ratio to check the measure of ЄJ.

b) N opposite sin K hypotenuse 7.2 cm ______MN M 7.8 cm ______sin K ______NK 7.2 K ______sin K ______7.8 ЄK Џ ______67° Find the measure of ЄK. The acute angles have a sum of 90°. MN is the side ______opposite ЄK. So, ЄN 90° ______ЄK NK is the ______hypotenuse . ЄN Џ 90° ______67° So, use the ______sine ratio. DRAFTЄN Џ ______23° We could use the cosine ratio to check the measure of ЄN.

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Example 2 Finding the Lengths of All Sides

Find all unknown side lengths to the nearest tenth of a metre.

P Q

76° 24.3 m R

Solution

side opposite ∠R Find the length of PR. P Q PR is the side adjacent to ЄR. side adjacent to ∠R 76° QR is the hypotenuse. 24.3 m R hypotenuse So, use the cosine ratio. adjacent cos R hypotenuse PR cos R Substitute: ЄR 76° and QR 24.3 QR PR cos 76° Multiply both sides by 24.3. 24.3 24.3 cos 76° PR PR 5.8787… PR is about 5.9 m long.

Find the length of PQ. PQ is the side opposite ЄR. QR is the hypotenuse. So, use the sine ratio. opposite sin R hypotenuse PQ sin R Substitute: ЄR 76° and QR 24.3 QR PQ DRAFTTEACHER NOTE sin 76° Multiply both sides by 24.3. 24.3 If students need to 24.3 sin 76° PQ review the Pythagorean We could use the Theorem, refer them to PQ 23.5781… Pythagorean Theorem to Chapter 1, Lesson 1.4 PQ is about 23.6 m long. check the side lengths. of this text.

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Check

1. Find all unknown side lengths to the nearest tenth of a metre.

18.4 m

43°

Find the length of ST. Find the length of SU. TU is the side ______opposite ЄS. TU is the side ______opposite ЄS. ST is the ______hypotenuse . SU is the side ______adjacent to ЄS. So, use the ______sine ratio. So, use the ______tangent ratio.

opposite opposite ______sin S ______hypotenuse ______tan S ______adjacent

TU TU ______sin S ______ST ______tan S ______SU 18.4 18.4 sin 43° tan 43° ST SU ST sin 43° 18.4 SU tan 43° 18.4 18.4 18.4 ST SU sin 43Њ tan 43Њ

ST ______26.9795… SU ______19.7315… ST is about 27.0 m long. SU is about 19.7 m long. ______DRAFT______

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Example 3 Solving a Triangle

Solve this triangle. B Give angle measures to the nearest degree. Give side lengths to the nearest tenth of a centimetre. 12.2 cm 19.2 cm

Solution C

Find the measure of ЄB. AB is the side adjacent to ЄB. We know the lengths BC is the hypotenuse. of two sides. So, use the cosine ratio. adjacent cos B hypotenuse AB cos B Substitute: AB 12.2 and BC 19.2 BC 12.2 cos B 19.2 ЄB 50.5491…° ЄB Џ 51° The acute angles in a right triangle have a sum of 90°. So, ЄC 90° ЄB ЄC Џ 90° 51° ЄC Џ 39°

Find the length of AC. Use the Pythagorean Theorem to find AC. 2 2 2 AC BC AB We could have used the AC2 19.22 12.22 sine ratio and the exact DRAFTmeasure of ЄB to find AC. AC2 219.8 AC Ί 219.8 AC 14.8256… AC is about 14.8 cm long.

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Check

1. Solve this triangle. Give side lengths to the nearest tenth of a centimetre.

E D We know the length of one side and the measure 17.3 cm of one acute angle. 62° F

The acute angles have a sum of ______90° . So, ЄE ______90° ______ЄF ЄE ______90° ______62° ЄE ______28° Find the length of DF. Find the length of DE. DF is the side ______adjacent to ЄF. DE is the side ______opposite ЄF. EF is the ______hypotenuse . EF is the ______hypotenuse . So, use the ______cosine ratio. So, use the ______sine ratio.

adjacent opposite ______cos F ______hypotenuse ______sin F ______hypotenuse

DF DE ______cos F ______EF ______sin F ______EF DF DE cos 62° sin 62° 17.3 17.3 17.3 cos 62° DF 17.3 sin 62° DE DF 8.1218… DE 15.2749… DF is about ______8.1 cm long. DE is about ______15.3 cm long. Practice DRAFT 1. Which ratio would you use to find the measure of each angle? a) ЄP b) ЄE Q D 4.8 cm Remember the acronym E SOH–CAH–TOA. 18.6 cm C 7.3 cm

P R 13.4 cm QR is the side ______opposite ЄP . DE is the ______side adjacent to ЄE . PR is the side ______adjacent to ЄP . CE is the ______hypotenuse . So, use the ______tangent ratio. So, use the ______cosine ratio.

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2. Which ratio would you use to find the length of each indicated side? a) GH b) MN 2.5 cm N H F 6.3 cm

51° M P 21° G

HF is the ______side opposite ЄG . MN is the ______side opposite ЄP . GH is the ______hypotenuse . NP is the ______side adjacent to ЄP . So, use the ______sine ratio. So, use the ______tangent ratio.

3. Find all unknown angle measures to the nearest degree. a) V b) Q R 4.2 cm 17.6 cm U 13.1 cm 5.5 cm W S

Find the measure of ЄU. Find the measure of ЄQ. opposite opposite ______sin U ______hypotenuse ______tan Q ______adjacent

VW RS ______sin U ______UW ______tan Q ______QS

4.2 13.1 ______sin U ______5.5 ______tan Q ______17.6

ЄU Џ ______50° ЄQ Џ ______37° The acute angles have a sum of 90°. The acute angles have a sum of 90°. So, ЄW 90° ______ЄU So, ЄR ______90° ЄQ ЄW Џ 90° ______50°DRAFTЄR Џ ______90° 37° ЄW Џ ______40° ЄR Џ ______53°

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4. Find all unknown side lengths to the nearest tenth of a centimetre. R

8.7 cm

48° S Q

Find the length of QS. Find the length of RS. opposite opposite ______sin S ______hypotenuse ______tan S ______adjacent

QR QR ______sin S ______QS ______tan S ______RS

8.7 8.7 sin 48° tan 48° QS RS

QS sin 48° 8.7 RS tan 48° 8.7

8.7 8.7 QS RS sin 48Њ tan 48Њ QS 11.7070… RS 7.8335… QS is about ______11.7 cm long. RS is about ______7.8 cm long.

5. Solve this triangle. Give angle measures to D 13.8 cm E the nearest degree. Give side lengths to the nearest tenth of a centimetre. 21.5 cm

C Find the measure of ЄE. Find the length of CD. adjacent Use the Pythagorean Theorem. ______cos E ______hypotenuse CD2 CE2 DE2 CD2 21.52 13.82 DE DRAFT 2 ______cos E ______CE CD 271.81 CD Ί 271.81 13.8 ______cos E ______21.5 CD 16.4866… CD is about ______16.5 cm long. ЄE Џ ______50° The acute angles have a sum of 90°. So, ЄC 90° ______ЄE ЄC Џ 90° ______50° ЄC Џ ______40°

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6. The base of a ladder is on level ground 1.9 m from a wall. The ladder leans against the wall. The angle between the ladder and the ground is 65°. a) How far up the wall does the ladder reach? b) How long is the ladder? Give your answers to the nearest tenth of a metre.

65° H 1.9 m G a) We want to find the length of FH. The measure of ЄG is known. FH is the side opposite ЄG. GH is the side adjacent to ЄG. So, use the tangent ratio. opposite ؍ tan G adjacent FH ؍ tan G GH FH ؍ tan 65° 1.9 FH ؍ tan 65° 1.9 …4.0745 ؍ FH The ladder reaches the wall at a height of about ______4.1 m .

b) We want to find the length of FG. GH is the side adjacent to ЄG. FG is the hypotenuse.DRAFT So, use the cosine ratio. adjacent ؍ cos G hypotenuse GH ؍ cos G FG 1.9 ؍ cos 65° FG TEACHER NOTE ؍ FG cos 65° 1.9 Next Steps: Have 1.9 students complete ؍ FG cos 65Њ questions 6, 7, 8, 12, and 13 on pages 111 ؍ FG 4.4957… and 112 of the The ladder is about ______4.5 m long. Student Text. 50 MMS10_L7.qxp 1/21/10 10:43 AM Page 51

2.7 Solving Problems Involving More than One Right Triangle

FOCUS Use trigonometric ratios to solve problems that involve more than one right triangle.

When a problem involves more than one right triangle, we can use information from one triangle to solve the other triangle.

Example 1 Solving a Problem with Two Triangles

Find the length of BC to the nearest tenth of a centimetre.

22.9 cm To solve a right triangle we must know: • the lengths of two sides, or ° ° • the length of one side and the A 26 49 C D measure of one acute angle

Solution

First use ᭝ABD to find the length of BD. opposite B sin A hypotenuse 22.9 cm Side BD is common to both triangles. BD sin A ° A 26 AB D BD sin 26° 22.9 22.9 sin 26° BD BD 10.0386… Do not clear the calculator screen.

In ᭝BCD, find the length of BC. opposite B sin C DRAFT hypotenuse 10.0386 cm BD sin C 49° BC D C 10.0386... sin 49° BC BC sin 49° 10.0386…

10.0386... BC sin 49 BC 13.3014… BC is about 13.3 cm long.

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Check

1. Find the measure of ЄF to the nearest degree. E D 34°

14.4 cm F G 9.6 cm

E Use ᭝DEG to find the length of EG. D 34° Use the sine ratio. Side EG is common to both triangles. 14.4 cm opposite sin D ______hypotenuse G

EG sin D ______DG EG sin ______34° ______14.4 E 14.4 sin 34° EG 8.0523 cm EG ______8.0523… F G 9.6 cm In ᭝EFG, use the ______tangent ratio to find ЄF.

opposite tan F TEACHER NOTE adjacent Show students how EG to use the ANS key to tan F FG find ЄF directly. 8.0523... is already 8.0523... on the calculator tan F 9.6 screen. Input: W683< Є F 39.9895…° %@%M< The measure of ЄF is aboutDRAFT ______40° . to display 39.98956614. The angle of elevation is the angle between object the horizontal and a person’s line of sight to an object above. angle of elevation horizontal

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Example 2 Solving a Problem Involving Angle of Elevation

Jason is lying on the ground midway between two trees, 100 m apart. The angles of elevation of the tops of the trees are 13° and 18°. How much taller is one tree than the other? Give the answer to the nearest tenth of a metre. N K 13° 18° Tree 1 M Tree 2 J P 100 m Solution

Jason is midway between the trees. 100 m So, the distance from Jason to the base of each tree is: 50 m 2 Use ᭝JKM to find the length of JK.

K 13° Є Tree 1 We know M 13°. JM JK is opposite ЄM. 50 m JM is adjacent to ЄM. opposite Use the tangent ratio. tan M adjacent JK tan M Substitute: ЄM 13° and JM 50 JM JK tan 13° 50 50 tan 13° JK JK 11.5434… Use ᭝MNP to find the length of NP. N 18° Tree 2 We know ЄM 18°. M P NP is opposite ЄM. 50 m MP is adjacent to ЄM. opposite Use the tangent ratio. tan M adjacent DRAFT NP tan M Substitute: ЄM 18° and MP 50 MP NP tan 18° 50 50 tan 18° NP NP 16.2459…

To find how much taller one tree is than the other, subtract: 16.2459 m 11.5434 m 4.7025 m One tree is about 4.7 m taller than the other.

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Check

1. The angle of elevation of the top of a tree, T, is 27°. H From the same point on the ground, the angle of elevation of a hawk, H, flying directly above the T

tree is 43°. The tree is 12.7 m tall. How high is 12.7 m the hawk above the ground? Give your answer 43° Q G to the nearest tenth of a metre. 27°

We want to find the length of HG. T Use ᭝QTG to find the length of QG. 12.7 m Use the tangent ratio. Q G 27° opposite tan Q ______adjacent TG tan Q Substitute: ______ЄQ 27° and ______TG 12.7 ______QG 12.7 tan ______27° ______QG QG tan 27° 12.7 12.7 QG tan 27Њ QG ______24.9251…

In ᭝QHG, use the tangent ratio to find HG. H opposite tan Q adjacent HG tan Q QG 43° Q G HG 24.9251 m tan 43° 24.9251... 24.9251… tan 43° HG HG ______DRAFT23.2430… The hawk is about ______23.2 m above the ground.

The angle of depression is the angle between horizontal the horizontal and a person’s line of sight

to an object below. angle of depression

object

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Example 3 Solving a Problem Involving Angle of Depression

From a small plane, V, the angle of depression of a sailboat is 21°. The angle of depression of a ferry on the other side of the plane is 52°. The plane is flying at an altitude of 1650 m. How far apart are the boats, to the nearest metre? V 21° 52° 1650 m

U XW sailboat ferry Solution

We want to find the length of UW. The angle of depression of the sailboat is 21°. So, in ᭝UVX, ЄV 90° 21°, or 69°.

V Use ᭝UVX to find the length of UX. Є 21° We know V 69°. 69° Є opposite 1650 m UX is opposite V. tan V Є adjacent VX is adjacent to V. U X So, use the tangent ratio. UX tan V Substitute: ЄV 69° and VX 1650 VX UX tan 69° 1650 1650 tan 69° UX UX 4298.3969…

The angle of depression of the ferry is 52°. So, ЄV in ᭝VWX is: 90° 52°, or 38°. V We know ЄV 38°. 38° 52° Use ᭝VWX to find the length of WX. WX is opposite ЄV. 1650 m VX is adjacent to ЄV. opposite DRAFT tan V So, use the tangent ratio. adjacent XW ferry WX tan V Substitute: ЄV 38° and VX 1650 VX WX tan 38° 1650 1650 tan 38° WX WX 1289.1212… To find the distance between the boats, add: 4298.3969 m 1289.1212 m 5587.5181 m The boats are about 5588 m apart.

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Check

1. This diagram shows a falcon, F, on a tree, with a squirrel, S, and a chipmunk, C, on the ground. From the falcon, the angles of depression of the animals are 36° and 47°. How far apart are the animals on the ground to the nearest tenth of a metre? 36° F 47°

15 m

GCS We want to find the length of CS. 36° F CS GS GC 54° The angle of depression of the squirrel is ______36° . So, ЄF in ᭝FSG is: 90° ______,36° or ______54° . 15 m

Use ᭝FSG to find the length of GS. G S opposite tan ____F ______adjacent GS tan ____F ______FG GS tan _____54° ______15 15 tan 54° GS GS ______20.6457… The angle of depression of the chipmunk is ______47° . So, ЄF in ᭝FCG is: 90° ______47° , or ______43° .

᭝ F Use FCG to find the length of GC. 47° opposite DRAFT43° tan F adjacent 15 m GC tan F FG GC GC tan 43° 15 15 tan 43° GC GC ______13.9877…

To find the distance between the animals, subtract: ______20.6457 m ______13.9877 m ______6.6580 m The animals on the ground are about ______6.7 m apart.

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Practice

B 1. Find the measure of ЄC to the nearest degree. C Use ᭝ABD to find the length of BD. 22.5 cm ° B A 37 12.6 cm D

37° A 12.6 cm D

Use the tangent ratio. opposite tan A ______adjacent

BD tan A ______AD BD tan ______37° ______12.6 12.6 tan 37° BD BD ______9.4947…

In ᭝BCD, use the ______sine ratio to find ЄC.

B C 9.4947 cm 22.5 cm

opposite sin C hypotenuse BD sin C TEACHER NOTE CD To find ЄC directly, DRAFT with 9.4947... already 9.4947 ... sin C 22.5 on the calculator screen, input: W//82< ЄC ______24.9603…° %>%M< The measure of ЄC is about ______25° . to display 24.96030534.

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2. Two guy wires support a flagpole, FH. The first wire is 11.2 m long and has an angle of inclination of 39°. The second wire has an angle of inclination of 47°. How tall is the flagpole to the nearest tenth of a metre?

G Recall that the angle the wire makes with the ground is called the angle of inclination. 11.2 m

39° 47° E H

We want to find the length of FH. G Use ᭝EGH to find the length of EH. Use the cosine ratio. 11.2 m adjacent ϭ cos E 39° ______hypotenuse E H EH ϭ cos E ______EG

EH cos ______39° ϭ ______11.2

EH ؍ cos 39° 11.2 EH ϭ ______8.7040…

In ᭝EFH, use the ______tangent ratio to find the length of FH. F opposite ؍ tan E adjacent FH ؍ tan E EH 47° FH E H ؍ tan 47° 8.7040 ... 8.7040 m FHDRAFT ؍ ؋ tan 47° …8.7040 FH ϭ ______9.3339… The flagpole is about ______9.3 m tall.

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3. A mountain climber is on top of a mountain that is 680 m high. The angles of depression of two points on opposite sides of the mountain are 48° and 32°. How long would a tunnel be that runs between the two points? Give your answer to the nearest metre. M 48° 32°

680 m

Q P N

We want to find the length of QN. M 48° The angle of depression of point Q is ______48° . 42° So, ЄM in ᭝PQM is: 90° ______48° , or ______42° . 680 m Use ᭝PQM to find the length of PQ. Q P Use the ______tangent ratio. opposite tan M adjacent PQ tan M MP PQ tan 42° 680 680 tan 42° PQ

PQ ______612.2747… The angle of depression of point N is ______32° . M 32° So, ЄM in ᭝PMN is: 90° ______32° , or ______58° . 58° 680 m Use ᭝PMN to find the length of PN. Use the tangent ratio. ______P N opposite tan M adjacent NP tan M MP NP DRAFT tan 58° 680 680 tan 58° NP NP ______1088.2274…

The length of the tunnel is: ______QN ______QP ______PN TEACHER NOTE Next Steps: Have QN 612.2747 m 1088.2274 m students complete questions 3, 4, 5, 6, 8, QN ______1700.5021 9, and 11 on pages 118 and 119 of the The tunnel would be about ______1701 m long. Student Text.

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Chapter 2 Puzzle

Angle Mania!

A. Find the angles of inclination of the diagonals shown. Assume the squares have side length 1 unit.

Sample Solution: A E 1 Diagonal PA: tan P ; ЄP 45° 1 unit 1 D PK 2 1 unit Diagonal PB: tan P ; ЄP Џ 63° 1 C Diagonal PC: tan P 3; ЄP Џ 72° Diagonal PD: tan P 4; ЄP Џ 76° B Diagonal PE: tan P 5; ЄP Џ 79° F AF G H J 1 unit 1 Є Џ Diagonal PF: tan P ; P 27° PM 2 2 units 1 P KM Diagonal PG: tan P ; ЄP Џ 18° 3 1 Diagonal PH: tan P ; ЄP Џ 14° 4 1 Diagonal PJ: tan P ; ЄP Џ 11° 5 B. How many squares would be needed on the vertical rectangle for a diagonal to have an angle of inclination greater than:

• 80°? tan 80° 5.6712… 6 squares • 88°? tan 88° 28.6362… 29 squares • 85°? tan 85° 11.4300… 12 squares • 89°? tan 89° 57.2899… 58 squares

C. How many squares would be needed on the horizontal rectangle 1 unit for a diagonal to have an angle of inclination less than: 1 ° adjacent • 10°? tan 10° 0.1763… 0.1763… 10 DRAFTadjacent 1 adjacent 0.1763… adjacent 5.6712… 6 squares 1 • 5°? tan 5° 0.0874… 11.4300… 12 squares 0.0874… 1 • 2°? tan 2° 0.0349… 28.6362… 29 squares TEACHER NOTE 0.0349 … Strategies may vary. 1 Some students may • 1°? tan 1° 0.0174… 57.2899… 58 squares use guess and test. 0.0174 …

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Chapter 2 Study Guide

Skill Description Example Find a trigonometric In ABC, B hypotenuse ratio. A B 10 6 side adjacent to ∠A side opposite ∠A

C A 8 C opposite opposite sin A sin A hypotenuse hypotenuse adjacent BC cos A sin A hypotenuse AB opposite 6 tan A sin A , or 0.6 adjacent 10 Find the measure of To find the measure of an acute angle To find the measure of B an angle. in a right triangle: in ABC above: 1. Use the given lengths to write a opposite tan B trigonometric ratio. adjacent 2. Use the inverse function on a AC tan B scientific calculator to find the BC measure of the angle. 8 tan B 6 8 B tan1 a6b B 53°

Find the length of To find the length of a side in a right To find the length of EF in a side. triangle: DEF: 1. Use the measure of an angle and E the length of a related side to write 64° anDRAFT equation using a trigonometric 3.0 cm ratio. D F 2. Solve the equation. adjacent cos E hypotenuse DE cos E EF 3.0 cos 64° EF EF cos 64° 3.0 3.0 EF cos 64 EF 6.8435… EF 6.8 cm

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Chapter 2 Review

2.1 1. Find the measure of ЄP to the nearest degree. Q opposite ϭ tan P ______adjacent

14 cm QR ϭ tan P ______PR 14 R tan P ϭ ______10 P 10 cm ЄP Џ ______54°

2.2 2. Find the length of TU to the nearest tenth of a centimetre.

T U opposite ϭ tan S ______adjacent 7.2 cm TU tan S ϭ ST 58° ______S TU tan ______58° ϭ ______7.2

TU ؍ tan 58° 7.2

TU ϭ ______11.5224…

TU is about ______11.5 cm long.

3. A flagpole casts a shadow that is 25 m long when the Z angle between the sun’s rays and the ground is 40°. What is the height of the flagpole to the nearest metre? ______ZX is the side oppositeDRAFT ЄY. Є 40° ______XY is the side adjacent to Y. Y X 25 m opposite ϭ tan Y ______adjacent ZX ϭ tan Y ______XY ZX ؍ tan 40° 25 ZX ؍ tan 40° 25 ZX ϭ ______20.9774… The flagpole is about ______21 m high.

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2.3 4. Use the information in the diagram to find the height of the tower of a wind turbine observed with a drinking-straw clinometer. Give the answer to the nearest tenth of a metre.

ЄA ϭ 90° Ϫ ______,17° or ______73° Side opposite ЄA: ______BC Side adjacent to ЄA: ______AC B opposite ϭ tan A ______adjacent

BC tan ______73° ϭ ______AC 180 160

BC 140 tan 73° 15 120

15 tan 73° BC 100 BC ϭ ______49.0627…

° 17 80

So, height of tower ϭ ______49.0627 m ϩ ______1.5 m

60 60 60

40 40

ϭ 50.5627 m 40 0 0 ______0 20 20 20 20 20 The height of the tower is about ______.50.6 m

A C 1.5 m 15 m This diagram is not drawn to scale.

2.4 5. Find the measure of each indicated angle to the nearest degree. a) D b) F 6.7 cm 13.6 cm 8.5 cm H 13.9 cm G

C E DRAFTHF is the ______.side adjacent to ЄH GH is the ______hypotenuse . CD is the ______.side opposite ЄE So, use the ______cosine ratio. DE is the ______hypotenuse . adjacent So, use the ______sine ratio. cos H hypotenuse opposite sin E HF hypotenuse cos H GH CD sin E 6.7 DE cos H 13.9 8.5 sin E ЄH Џ ______61° 13.6 ЄE Џ ______39°

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6. A 2.8-m ladder is leaning against a barn, as shown. What angle does the ladder make with the barn? Give your answer to the nearest degree. J We want to find the measure of _____J . KM is the ______side opposite ЄJ . 2.8 m JK is the ______hypotenuse . So, use the ______sine ratio.

opposite M K sin J 1.3 m hypotenuse KM sin J JK Assume the ground is horizontal. 1.3 sin J 2.8 J ______28° The angle the ladder makes with the barn is about ______28° .

2.5 7. Find the length of each indicated side to the nearest tenth of a centimetre. a) RS b) NQ T P

57° 8.8 cm 17.2 cm 24° N Q S The measure of N is known. R Use the ______cosine ratio. adjacent The measure of T is known. cos N Use the ______sine ratio. hypotenuse opposite NP sin T cos N hypotenuse NQ 8.8 RS cos 24° sin T NQ TR DRAFT NQ cos 24° 8.8 sin 57° RS 17.2 8.8 NQ cos 24Њ 17.2 sin 57° RS RS 14.4251… NQ 9.6327… RS is about ______14.4 cm long. NQ is about ______9.6 cm long.

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8. An escalator is 14.5 m long. The escalator makes an angle of 27° with the ground. What is the height of the escalator? Give your answer to the nearest tenth of a metre.

14.5 m

27° A C To find the length of BC, use the ______sine ratio. opposite sin A hypotenuse BC sin A AB BC sin 27° 14.5 14.5 sin 27° BC BC ______6.5828… The escalator is about ______6.6 m high.

2.6 9. Solve this triangle. Give side lengths to the nearest tenth of a centimetre.

8.5 cm

° E 51 G

Find the length of FG. Find the length of EG. Use the ______tangent ratio. Use the ______cosine ratio. opposite adjacent tan E cos E adjacent hypotenuse FG EF tan E cos E EF DRAFTEG FG 8.5 tan 51° cos 51° 8.5 EG 8.5 tan 51° FG EG cos 51° 8.5 8.5 FG 10.4966… EG cos 51Њ FG is about ______10.5 cm long. EG 13.5066… The acute angles have a sum of ____90° . EG is about ______13.5 cm long.

So, G 90° ______ЄE G 90° ______51° G ______39°

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2.7 10. Two buildings are 25 m apart. From the top of the shorter building, the angles of elevation and depression of the top and bottom of the taller building are 31° and 48° respectively. What is the height of the taller building? Give your answer to the nearest metre. G

31° F H 48° Taller building Shorter building

K 25 m J

25 m We want to find the length of GJ. F H 48° GJ ______HJ ______GH The angle of depression of point J is ______48° . Use FHJ to find the length of HJ. Use the ______tangent ratio. opposite J tan F adjacent HJ tan F FH HJ tan 48° 25 25 tan 48° HJ HJ ______27.7653… The angle of elevation of point G is ______31° . G Use FGH to find the length of GH. tangent Use the ______ratio. 31° F H opposite 25 m tan F adjacent GH tan 31° 25 DRAFT 25 tan 31° GH TEACHER NOTE GH ______15.0215… Next Steps: Direct students to questions To find the height of the taller building, add: 1, 6, 8, 13, 15, 17, 19, ______27.7653 m ______15.0215 m ______42.7868 m 22, and 23 on pages 124–126 of the The taller building is about ______43 m tall. Student Text.