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48

Chapter 2 Right Triangle Trigonometry

2.1 Trigonometric Functions of an Acute Angle

2.1 Practice Exercises

1.

hypotenuse 5csc

opposite 3hypotenuse 5

secadjacent 4

adjacent 4cot

opposite 3hypotenuse 5

cscopposite 4

hypotenuse 5sec

adjacent 3adjacent 3

cotopposite 4

A

A

A

B

B

B

= =

= =

= =

= =

= =

= =

2. a. ( )cos 43 sin 90 43 sin 47° = ° − ° = °

b. ( )cot 70 tan 90 70 tan 20° = ° − ° = °

c. ( )sec54 csc 90 54 csc36° = ° − ° = °

3. Because the cosine and sine are cofunctions, cos A = sin B if A and B are acute angles and A + B = 90°. Thus,

( ) ( )cos 40 sin 40 902 50 25

θ θ θ θθ θ

+ ° = ⇒ + ° + = °⇒= °⇒ = °

4.

First use the Pythagorean theorem to find b. 2 2 2 2 2 2 21 5 24

24 2 6

a b c b b

b

+ = ⇒ + = ⇒ = ⇒= =

Now use a = 1 = opposite, 2 6b = = adjacent, and c = 13 = hypotenuse.

1sin csc 5

52 6 5 5 6

cos sec5 122 61 6

tan122 6

cot 2 6

a cA A

c ab a

A Ac ca

Abb

Aa

= = = =

= = = = =

= = =

= =

5.

The distance that the boat is from the dock is represented by the side opposite the 60° angle. We are given the adjacent side, so we can use the tangent function. Let x = the length of the opposite side.

tan 60 5 tan 60 5 3 8.75

xx° = ⇒ = ° = ≈

The boat is approximately 8.7 feet from the dock.

6. a.

b.

7. a.

b.

http://www.parmispdf.com

Section 2.1 Trigonometric Functions of an Acute Angle 49

8.

We have the opposite side and the hypotenuse, so use the sine function.

5sin 0.25

20θ = =

14.5θ ≈ °

2.1 A Exercises: Basic Skills and Concepts

1. In a right triangle ABC with right angle at C, opposite adjacent

sin , cos ,hypotenuse hypotenuse

A A= = and

oppositetan .

adjacentA =

2. If θ and α are two acute angles (measured in degrees) in a right triangle and cos sin ,θ α= then 90 .θ α+ = °

3. To find the value of csc 63° on a scientific calculator, we find sin 63° and then use the

1x− key.

4. False. In order to determine the sine and cosine, we must know the hypotenuse.

5. True

6. True

For exercises 7–12, recall that opposite hypotenuse

sin cschypotenuse opposite

adjacent hypotenusecos sec

hypotenuse adjacentopposite adjacent

tan cotadjacent opposite

θ θ

θ θ

θ θ

= =

= =

= =

7.

2 2 5 5 5sin csc

25 25 511 11 5 5 5

cos sec25 115 5

2 11tan cot

11 2

θ θ

θ θ

θ θ

= = =

= = =

= =

8.

7 2 7 2sin csc 2

2 77 27 2 7 2

cos sec 22 77 2

7 7tan 1 cot 1

7 7

θ θ

θ θ

θ θ

= = = =

= = = =

= = = =

9.

6 3 10 5sin csc

10 5 6 38 4 10 5

cos sec10 5 8 46 3 8 4

tan cot8 4 6 3

θ θ

θ θ

θ θ

= = = =

= = = =

= = = =


50 Chapter 2 Right Triangle Trigonometry

10.

8 17sin csc

17 815 17

cos sec17 158 15

tan cot15 8

θ θ

θ θ

θ θ

= =

= =

= =

11.

9 41sin csc

41 940 41

cos sec41 409 40

tan cot40 9

θ θ

θ θ

θ θ

= =

= =

= =

12.

4 1 12sin csc 3

12 3 48 2 2 2 12 3 2

cos sec12 3 48 24 2 8 2

tan cot 2 24 48 2

θ θ

θ θ

θ θ

= = = =

= = = =

= = = =

13. ( )sin 58 cos 90 58 cos 32° = ° − ° = °

14. ( )cos 37 sin 90 37 sin 53° = ° − ° = °

15. ( )tan 27 cot 90 27 cot 63° = ° − ° = °

16. ( )cot 49 tan 90 49 tan 41° = ° − ° = °

17. ( )sec 65 csc 90 65 csc 25° = ° − ° = °

18. ( )csc 78 sec 90 78 sec12° = ° − ° = °

19. ( )sin cos 30 30 902 60 30

θ θ θ θθ θ

= + ° ⇒ + + ° = °⇒= °⇒ = °

20. ( )cot tan 10 10 902 80 40

θ θ θ θθ θ

= + ° ⇒ + + ° = °⇒= °⇒ = °

21. ( )sec csc 30 30 902 120 60

θ θ θ θθ θ

= − ° ⇒ + − ° = °⇒= °⇒ = °

22. ( )cos sin 10 10 902 100 50

θ θ θ θθ θ

= − ° ⇒ + − ° = °⇒= °⇒ = °

23. ( ) ( )tan 5 cot 155 15 90

2 20 90 2 70 35

θ θθ θθ θ θ

+ ° = + ° ⇒+ ° + + ° = °⇒+ ° = °⇒ = °⇒ = °

24. ( ) ( )csc 8 sec 328 32 90

2 40 90 2 50 25

θ θθ θθ θ θ

+ ° = + ° ⇒+ ° + + ° = °⇒+ ° = °⇒ = °⇒ = °

25. ( )sin 3 cos 30 3 30 904 60 15

θ θ θ θθ θ

= + ° ⇒ + + ° = °⇒= °⇒ = °

26. ( ) ( )cot 4 2 tan 434 2 43 905 45 90 5 45 9

θ θθ θθ θ θ

+ ° = + ° ⇒+ ° + + ° = °⇒+ ° = °⇒ = °⇒ = °

For exercises 27–32, first find the length of the missing side using the Pythagorean theorem, then find the trigonometric functions of A using

sin csc

cos sec

tan cot

a cA A

c ab c

A Ac ba b

A Ab a

= =

= =

= =

27. 2 2 2 2 2 29 12 225 15c a b c c= + ⇒ = + = ⇒ = 9 3 15 5

sin csc15 5 9 312 4 15 5

cos sec15 5 12 49 3 12 4

tan cot12 4 9 3

A A

A A

A A

= = = =

= = = =

= = = =

28. ( )22 2 2 2 23 7 16 4c a b c c= + ⇒ = + = ⇒ =

3 4sin csc

4 37 4 4 7

cos sec4 773 3 7 7

tan cot7 37

A A

A A

A A

= =

= = =

= = =



29. 2 2 2 2 2 2

2 2 229 21

29 21 400 20

c a b b

b b

= + ⇒ = + ⇒= − = ⇒ =

21 29sin csc

29 2120 29

cos sec29 2021 20

tan cot20 21

A A

A A

A A

= =

= =

= =

30. 2 2 2 2 2 2

2 2 25 2

5 22 21 21

c a b b

b b

= + ⇒ = + ⇒= − = ⇒ =

2 5sin csc

5 221 5 5 21

cos sec5 21212 2 21 21

tan cot21 221

A A

A A

A A

= =

= = =

= = =

31. ( )( )

22 2 2 2 2

22 2

7 13

7 13 36 6

c a b a

a a

= + ⇒ = + ⇒

= − = ⇒ =

6 7sin csc

7 613 7 7 13

cos sec7 13136 6 13 13

tan cot13 613

A A

A A

A A

= =

= = =

= = =

32. ( )( )

22 2 2 2 2

22 2

5 1

5 1 4 2

c a b a

a a

= + ⇒ = + ⇒

= − = ⇒ =

2 2 5 5sin csc

5 251 5 5

cos sec 55 15

2 1tan 2 cot

1 2

A A

A A

A A

= = =

= = = =

= = =

In exercises 33–42, round each answer to four decimal places.

33.

0.5110≈

34.

7.2848≈

35.

1.1371≈

36.

20.8188

37.

0.6552≈

38.

0.9550≈

39.

1.1294≈

40.

0.7501≈

41.

2.8239≈



42.

1.0051≈

43. 1

cos11.8 0.9789sec11.8

= ° ≈°

44. 1

tan 55.8 1.4715cot 55.8

= ° ≈°

45. ( )cot 90 31.16 tan 31.16 0.6047° − ° = ° ≈

46. ( )cot 80 tan 90 80 tan10 0.1763° = ° − ° = ° ≈

47.

23.6θ ≈ °

48.

41.8θ = °

49.

58.1θ ≈ °

50.

78.5θ = °

51.

36.7θ = °

52.

55.5θ = °

53.

32.6θ = °

54.

64.8θ = °

2.1 B Exercises: Applying the Concepts

In exercises 55–60, draw a simple diagram to help organize the information in the problem.

55.

tan 45 1 4.$ 4

AC ACAC° = ⇒ = ⇒ =

The rope is anchored 4 feet from the base of the tree.

56.

sin 8 15sin 8 2.1.15

BCBC° = ⇒ = ° ≈

The ramp is about 2.1 feet above the ground.



57.

sin 40 30sin 40 19.30

BCBC° = ⇒ = ° ≈

The wire is attached about 19 feet up the flagpole.

58.

100 100

tan 40 119.tan 40

BCBC

° = ⇒ = ≈°

The tree was about 119 feet tall.

59.

35

tan tan 1.75 60 .20

θ θ θ= ⇒ = ⇒ ≈ °

The angle formed by the light beam and a vertical line through the spotlight is approximately 60°.

60.

40

tan tan 2.2222 66 .18

θ θ θ= ⇒ ≈ ⇒ ≈ °

The angle formed by the camera’s line of sight and a vertical line is approximately 66°.

2.1 C Exercises: Beyond the Basics

61.

In a cube, the length of each edge is equal, so

AB = BC = CD. In triangle ABC, angle B is a

right angle. Therefore, 2.AC AB= In triangle

ACD, 1 2

tan .22 2

DC AB

AC ABθ = = = =

62.

ABEF is a square, so AB = BE = EF = FA. We

are told that the length of the rectangle is twice the length of a side of the square, so AC = 2AB. The width of the rectangle is the same as the length of a side of the square, so CD = AB.

1tan 26.6

2 2

CD AB

AC ABθ θ= = = ⇒ ≈ °

63.

Triangle ABD is a 30°-60°-90° triangle, so

a = 2 and 2 3.b = Angle DBA measures 60°, so angle DBC measures 30°, and triangle BCD is also a 30°-60°-90° triangle. Since a = 2,

c = 1, and 3.d =

64.

Triangle ACD is a 30°-60°-90° triangle, so

10 10 3 20 3and .

3 33a b= = =

(continued on next page)



(continued from page 53)

Triangle BCD is also a 30°-60°-90° triangle, so

20 3 3 20.

33c = =

20 402 10 2 10

3 3103

AB c d d

d

⎛ ⎞= ⇒ + = ⇒ + = ⇒⎜ ⎟⎝ ⎠

=

65.

All radii of a circle are equal, so CD = CB.

Since AC = AC, ABC ADC≅△ △ by hypotenuse-leg. Therefore,

30DAC BAC∠ = ∠ = ° since corresponding parts of congruent triangles are equal. Therefore, triangle ABC is a 30°-60°-90° triangle, and AC = 2BC = 4 units.

66.

We want sin and cos , so 1 unit.a b cθ θ= = =

67.

In a triangle the length of side opposite the

smaller angle is shorter than the side opposite the larger angle. So, b > a. We have

sin sina

a cc

α α= ⇒ = and sinb

cβ = ⇒

sin .b c β= Therefore, b a> ⇒

sin sin sin sin .c cβ α β α> ⇒ > Similarly,

cos cosb

b cc

α α= ⇒ = and

cos cos .a

a cc

β β= ⇒ = Therefore, b a> ⇒

cos cos cos cos .c cα β α β> ⇒ >

68.

We want tan , so 1 unit.a bθ= =

69.

In a triangle the length of side opposite the

smaller angle is shorter than the side opposite the larger angle. So, b > a. We have

tan tana

a bb

α α= ⇒ = and tanb

aβ = ⇒

tan .b a β= Therefore, b a> ⇒

tan tan tan tan

tan tan tan tan 1.

a ba b

b bβ α β α

α β α β

> ⇒ > ⇒

> ⇒ >

If we divide both sides of tan tana bβ α> by

a, we obtain tan tana b

a aβ α> ⇒

tan tan tan 1 tan .β β α α> ⇒ > Since

tan 1 tan , tan tan .β α β α> > >

70.

Since triangle ABC is a right triangle,

A = 90° – B. So, ( )tan tan 90 cot .A B B= ° − =

2

tan and cot , so

.

CL LBA B

AL CLCL LB

CL AL LBAL CL

= =

= ⇒ = ⋅



2.1 Critical Thinking

71.

( ) ( )

( )

2 22 2

2 2 2 2 2 2

2 2 2 2 2

sin sin

cos cos

sin cos

sin cos

sin cos

aa c

cb

b cc

a b c c

a b c c

a b c

θ θ

θ θ

θ θθ θθ θ

= ⇒ =

= ⇒ =

+ = + ⇒+ = + ⇒+ = +

The Pythagorean theorem gives us 2 2 2,a b c+ = so we have

( )2 2 2 2

2 2

sin cos

sin cos 1.

c c θ θ

θ θ

= + ⇒

+ =

72.

In the right triangle OAB, tan ,mx

mx

θ = =

which is the slope of the given line.

2.1 Group Project

(i) In triangle ABC, angle ABC = 60° and angle ACB = 90°, so angle CAB = 30°. Triangle ABC

is a 30°-60°-90° triangle, so 3.b =

(ii) Since angle DAC = 45° and angle D = 90°, angle ACD = 45°. Thus, triangle ADC is an

isosceles right triangle with hypotenuse

3,b = so 3 6

.22

AD DC= = =

(iii) From (ii), we know that angle ACD = 45°. 180

45 90 180 45ACD ACB BCF

BCF BCF∠ + ∠ +∠ = °⇒° + ° + ∠ = °⇒ ∠ = °

Since 45BCF∠ = ° , FBC∠ also measures 45°. Therefore, triangle BCF is an isosceles right triangle with hypotenuse 1, so

1 2.

22CF BF= = =

(iv) From (ii), we know that 6

.2

DC = From (iii),

we know that 2

.2

CF = AE = DF since AEFD

is a rectangle, so

6 2 6 2

2 2 2AE DF DC CF

+= = + = + =

From (ii), we know that 6

,2

AD = and from

(iii), we know that 2

.2

BF = AD = EF since

AEFD is a rectangle, so

6 2 6 2.

2 2 2

BE EF BF AD BF

BE

= − = − ⇒−= − =

(vi) From (i), we know that angle CAB = 30°. 90

45 30 90 15DAC CAB BAE

BAE BAE∠ +∠ + ∠ = ° ⇒° + ° + ∠ = °⇒ ∠ = °

( )( )

180180 15 90 75

ABE BAE E∠ = ° − ∠ + ∠= ° − ° + ° = °

(vii)

6 22sin sin15

2 26 2

0.25884

6 2

2cos cos152 2

6 20.9659

4

BEBAE

AEBAE

−

= ° = =

−= ≈

+

= ° = =

+= ≈



(viii)

6 22sin sin 75

2 26 2

0.96594

6 2

2cos cos 752 2

6 20.2588

4

AEABE

BEABE

+

= ° = =

+= ≈

−

= ° = =

−= ≈

2.2 Solving Right Triangles

2.2 Practice Problems

1.

90 49.2 40.8A = ° − ° = °

sin sin 40.816.9

16.9sin 40.8 11.0 in.

cos 40.8 cos 40.816.9

16.9cos 40.8 12.8 in.

a aA

ca

b b

cb

= ⇒ ° = ⇒

= ° ≈

° = ⇒ ° = ⇒

= ° ≈

2.

2 2 2 2 2

7.72tan tan 0.8092

9.5439.090 39.0 51.0

9.54 7.72 150.61

150.61 12.3 in.

aA A

bA

B

c a b

c

= ⇒ = ≈ ⇒

≈ °= ° − ° = °

= + = + = ⇒= ≈

3.

From example 3, we know that x = 36 ft.

tan 50 36 tan 50 43 ft36

hh= ⇒ = ° ≈

4.

( )

sin 27 25sin 27 sin 2725

25sin 27 sin 27

25sin 2725sin 27 1 sin 27 21

1 sin 27

rr r

rr r

r r

° = ⇒ ° + ° = ⇒+

° = − °⇒°° = − ° ⇒ = ≈

− °

5.

425 425

tan 4.2 0.734

5787.4 ft or 1.096 mib b

b

° = ⇒ ≈ ⇒

≈

6.

sin 25 3.3387sin 253.3387

1.4110 mi 7450 ft

hh

h

° = ⇒ = °⇒

≈ ≈

The height of Mt. McKinley is about 12,870 + 7450 = 20,320 ft


Section 2.2 Solving Right Triangles 57

2.2A Exercises: Basic Skills and Concepts

1. The longest golf drive in competition was made by Mike Austin in 1974, and it was 515 yards. The range of values that would be reported as 515 yards is between 514.5 yards and 515.5 yards.

2. When solving a right triangle with side lengths given to three significant digits, the angles found should be given to the nearest 10 minutes or tenth of a degree.

3. If one acute angle in a right triangle is 26.4°, the other acute angle measures 63.6°.

4. If the hypotenuse of a right triangle measures 10 feet and one side measures 8 feet, then the remaining side measures 6 feet.

5. False. There is no uncertainty, so we don’t have to be concerned with rounding. (See page 67 in the text.)

6. True

7. A = 34°, c = 15 in. 90 34 56

sin sin 3415

15sin 34 8.4 in.

cos cos 3415

15cos34 12 in.

Ba a

Ac

ab b

Ac

b

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈

8. B = 63°, c = 18 in. 90 63 27

sin sin 2718

18sin 27 8.2 in.

cos cos 2718

18cos 27 16 in.

Aa a

Ac

ab b

Ac

b

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈

9. B = 52.37°, b = 11.41 90 52.37 37.63

tan tan 37.6311.41

11.41tan 37.63 8.796

11.41cos cos37.63

11.4114.41 in.

cos 37.63

A

a aA

ba

bA

c c

c

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ≈°

10. A = 47.38°, a = 29.14 90 47.38 42.62B = ° − ° = °

29.14tan tan 47.38

29.1426.81

tan 47.3829.14

sin sin 47.38

29.1439.60

sin 47.38

aA

b b

b

aA

c c

c

= ⇒ ° = ⇒

= ≈°

= ⇒ ° = ⇒

= ≈°

11. a = 15.47, b = 33.54 2 2 2 2 2

2 2

1

15.47 33.54

15.47 33.54 36.9415.47

tan33.54

15.47tan 24.76

33.5490 24.76 65.24

c a b

ca

Ab

A

B

−

= + = + ⇒

= + ≈

= = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °

12. a = 31.29, b = 21.16 2 2 2 2 2

2 2

1

31.29 21.16

31.29 21.16 37.7731.29

tan21.16

31.29tan 55.93

21.1690 55.93 34.07

c a b

ca

Ab

A

B

−

= + = + ⇒

= + ≈

= = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °

Use the triangle below to help you identify the opposite and adjacent legs.

13. A = 23.7°, c = 2.33 cm 90 23.7 66.3

sin sin 23.72.33

2.33sin 23.7 0.937 cm

cos cos 23.72.33

2.33cos 23.7 2.13 cm

Ba a

Ac

ab b

Ac

b

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈

14. A = 73.3°, c = 7.24 cm 90 73.3 16.7

sin sin 73.37.24

7.24sin 73.3 6.93 cm

cos cos 73.37.24

7.24cos 73.3 2.08 cm

Ba a

Ac

ab b

Ac

b

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈



In exercises 15–18, one of the given measurements has three significant digits, while the other has four. The answer should have three significant digits.

15. B = 32.6°, c = 64.21 ft 90 32.6 57.4

sin sin 32.664.21

64.21sin 32.6 34.6 ft

cos cos 32.664.21

64.21cos 32.6 54.1 ft

Ab b

Bc

ba a

Bc

a

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈

16. B = 37.6°, c = 14.42 ft 90 37.6 52.4

sin sin 37.614.42

14.42sin 37.6 8.80 ft

cos cos 37.614.42

14.42cos 37.6 11.4 ft

Ab b

Bc

ba a

Bc

a

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈

17. A = 62.92°, b = 14.7 ft 90 62.92 27.08

tan tan 62.9214.7

14.7 tan 62.92 28.8 ft

14.7cos cos 62.92

14.732.3 ft

cos 62.92

B

a aA

ba

bA

c c

c

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ≈°

18. A = 41.12°, b = 27.4 ft 90 41.12 48.88

tan tan 41.1227.4

27.4 tan 41.12 23.9 ft

27.4cos cos 41.12

27.436.4 ft

cos 41.12

B

a aA

ba

bA

c c

c

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ≈°

19. B = 28.47°, a = 5.243 m 90 28.47 61.53

tan tan 28.475.243

5.243 tan 28.47 2.843 m

5.243cos cos 28.47

5.2435.964 m

cos 28.47

A

b bB

ab

aB

c c

c

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ≈°

20. B = 71.43°, a = 22.14 m 90 71.43 18.57

tan tan 71.4322.14

22.14 tan 71.43 65.90 m

22.14cos cos 71.43

22.1469.52 m

cos 71.43

A

b bB

ab

aB

c c

c

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ≈°

21. A = 71.37°, a = 42.66 cm 90 71.37 18.63

42.66tan tan 71.37

42.6614.38 cm

tan 71.3742.66

sin sin 71.37

42.6645.02 ft

sin 71.37

B

aA

b b

b

aA

c c

c

= ° − ° = °

= ⇒ ° = ⇒

= ≈°

= ⇒ ° = ⇒

= ≈°

22. B = 61.42°, b = 27.41 cm 90 61.42 28.58

27.41tan tan 61.42

27.4114.93 cm

tan 61.4227.41

sin sin 61.42

27.4131.21 cm

sin 61.42

A

bB

a a

a

bB

c c

c

= ° − ° = °

= ⇒ ° = ⇒

= ≈°

= ⇒ ° = ⇒

= ≈°

23. a = 15.6 cm, b = 12.4 cm 2 2 2 2

1

15.6 12.4 19.9 cm15.6

tan tan12.4

15.6tan 51.5

12.490 51.5 38.5

c a ba

A Ab

A

B

−

= + = + ≈

= ⇒ = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °

24. a = 210 cm, b = 514 cm Note that there are only two significant digits in 210, so there should be two significant digits in the answers.

2 2 2 2

1

210 514 560 cm210

tan tan514

210tan 22

51490 22 68

c a ba

A Ab

A

B

−

= + = + ≈

= ⇒ = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °



In exercises 25–27, one of the given measurements has three significant digits, while the other has four. The answer should have three significant digits.

25. a = 5.71 m, b = 18.39 m 2 2 2 2

1

5.71 18.39 19.3 m5.71

tan tan18.39

5.71tan 17.2

18.3990 17.2 72.8

c a ba

A Ab

A

B

−

= + = + ≈

= ⇒ = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °

26. b = 60.4 m, c = 192.8 m 2 2 2 2 2 2

2 2 2

2 2

1

192.8 60.4

192.8 60.4

192.8 60.4 183 m60.4

cos cos192.8

60.4cos 71.7

192.890 71.7 18.3

c a b a

a

ab

A Ac

A

B

−

= + ⇒ = + ⇒= − ⇒

= − ≈

= ⇒ = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °

27. 34 7 , 5.278 ft

90 34 7 89 60 34 7 55 53

sin sin 34 75.278

5.278sin 34 7 2.960 ft

cos cos 34 75.278

5.278cos 34 7 4.370 ft

A c

B

a aA

ca

b bA

cb

= ° =′= ° − ° = ° − ° = °′ ′ ′ ′

= ⇒ ° = ⇒′

= ° ≈′

= ⇒ ° = ⇒′

= ° ≈′

28. 75 26 , 12.38 ft

90 75 26 89 60 75 26 14 34

sin sin 75 2612.38

12.38sin 75 26 11.98 ft

cos cos 75 2612.38

12.38cos 75 26 3.114 ft

A c

B

a aA

ca

b bA

cb

= ° =′= ° − ° = ° − ° = °′ ′ ′ ′

= ⇒ ° = ⇒′

= ° ≈′

= ⇒ ° = ⇒′

= ° ≈′

Use the figure below for exercises 29–36.

29. 37 , 49 , 17A BDC a= ° ∠ = ° =

In right triangle BDC, tanh

BDCx

∠ = ⇒

tan 49 tan 49 .h

h xx

° = ⇒ = °

In right triangle ABC, tanh

Aa x

∠ = ⇒+

( )tan 37 17 tan 37 .17

hh x

x° = ⇒ = + °

+ Now

equate the two expressions for h, and solve for x. ( )tan 49 17 tan 37x x° = + °⇒

( )

tan 49 tan 37 17 tan 37

tan 49 tan 37 17 tan 37

tan 49 tan 37 17 tan 37

17 tan 3732

tan 49 tan 37tan 49 32 tan 49 37

x x

x x

x

x

h x

° = ° + °⇒° − ° = °⇒° − ° = °⇒

°= ≈° − °

= ° = ° ≈

30. 21 , 53 , 8A BDC a= ° ∠ = ° =


BDCx

∠ = ⇒

tan 53 tan 53 .h

h xx

° = ⇒ = °


Aa x

∠ = ⇒+

( )tan 21 8 tan 21 .8

hh x

x° = ⇒ = + °

+ Now

equate the two expressions for h, and solve for x. ( )tan 53 8 tan 21x x° = + °⇒

( )

tan 53 tan 21 8 tan 21



8 tan 213

tan 53 tan 21tan 53 3 tan 53 4

x x

x x

x

x

h x

° = ° + °⇒° − ° = °⇒° − ° = °⇒

°= ≈° − °

= ° = ° ≈

31. 127 , 54 , 33A BDC c= ° ∠ = ° =

In right triangle ABC, 1

sinh

Ac

= ⇒

sin 27 33sin 27 15.33

hh° = ⇒ = ° ≈


BDCx

∠ = ⇒

15 15tan 54 11.

tan 54x

x° = ⇒ = ≈

°






cosx a

Ac

+= ⇒

11cos 27 33cos 27 11

3333cos 27 11 18.

aa

a

+° = ⇒ ° = + ⇒

= ° − ≈

32. 114 , 31 , 50A BDC c= ° ∠ = ° =


sinh

Ac

= ⇒

sin14 50sin14 12.50

hh° = ⇒ = ° ≈


BDCx

∠ = ⇒

12 12tan 31 20.

tan 31x

x° = ⇒ = ≈

°


Aa x

= ⇒+

12tan14 tan14 20 tan14 12

20tan14 12 20 tan14

12 20 tan1428.

tan14

aa

a

a

° = ⇒ ° + ° = ⇒+

° = − °⇒− °= ≈

°

33. 239 , 55 , 22A BDC c= ° ∠ = ° =

In right triangle BDC, 2

sinh

BDCc

∠ = ⇒

sin 55 22sin 55 18.22

hh° = ⇒ = ° ≈

In the same triangle, 2

cosx

BDCc

∠ = ⇒

cos 55 22cos 55 13.22

xx° = ⇒ = ° ≈


Aa x

= ⇒+

18tan 39 tan 39 13 tan 39 18

13tan 39 18 13 tan 39

18 13 tan 399.2

tan 39

aa

a

a

° = ⇒ ° + ° = ⇒+

° = − °⇒− °= ≈

°

34. 228 , 36 , 19A BDC c= ° ∠ = ° =


sinh

BDCc

∠ = ⇒

sin 36 19sin 36 11.19

hh° = ⇒ = ° ≈


cosx

BDCc

∠ = ⇒

cos 36 19cos 36 15.22

xx° = ⇒ = ° ≈


Aa x

= ⇒+

11tan 28 tan 28 15 tan 28 11

15a

a° = ⇒ ° + ° = ⇒

+

tan 28 11 15 tan 2811 15 tan 28

5.7tan 28

a

a

° = − °⇒− °= ≈

°

35. 13 , 25 , 11A DBC a= ° ∠ = ° =

In right triangle BDC, tanx

DBCh

∠ = ⇒

tan 25 tan 25 .x

x hh

° = ⇒ = °


Ax a

∠ = ⇒+

tan13 tan13 11tan1311

11tan13tan13 11tan13 .

tan13

hx h

xh

x h x

° = ⇒ ° + ° = ⇒+

− °° = − °⇒ =°

Equate the expressions for x, and solve for h.

( )

11tan13tan 25

tan13tan 25 tan13 11tan13

tan 25 tan13 11tan13

tan 25 tan13 1 11tan13

11tan132.8

tan 25 tan13 1

hh

h h

h h

h

h

− °° = ⇒°

° ° = − ° ⇒° ° − = − °⇒° ° − = − °⇒− °= ≈

° ° −

tan 25 2.8 tan 25 1.3x h= ° ≈ ° ≈

36. 24 , 31 , 40A DBC a= ° ∠ = ° =


DBCh

∠ = ⇒

tan 31 tan 31 .x

x hh

° = ⇒ = °


Ax a

∠ = ⇒+

tan 24 tan 24 40 tan 2440

40 tan 24tan 24 40 tan 24 .

tan 24

hx h

xh

x h x

° = ⇒ ° + ° = ⇒+

− °° = − ° ⇒ =°





Equate the expressions for x, and solve for h.

( )

40 tan 24tan 31

tan 24tan 31 tan 24 40 tan 24

tan 31 tan 24 40 tan 24

tan 31 tan 24 1 40 tan 24

40 tan 2424

tan 31 tan 24 1

hh

h h

h h

h

h

− °° = ⇒°

° ° = − °⇒° ° − = − °⇒° ° − = − °⇒− °= ≈

° ° −

tan 31 24 tan 31 14x h= ° ≈ ° ≈


37. 35 , 12A r= ° =

90 35 55

12sin 35 sin 35 12sin 35 12

12sin 35 12 12sin 35

12 12sin 358.9

sin 35

B

xx

x

x

= ° − ° = °

° = ⇒ ° + ° = ⇒+

° = − °⇒− °= ≈

°

38. 29 , 18B r= ° =

90 29 61

18cos 29

18cos 29 18cos 29 18

cos 29 18 18cos 29

18 18cos 292.6

cos 29

A

xx

x

x

= ° − ° = °

° = ⇒+

° + ° = ⇒° = − °⇒− °= ≈

°

39. 41 , 24A x= ° =

( )

90 41 49

sin 41 24sin 41 sin 4124

24sin 41 sin 41

24sin 4124sin 41 1 sin 41

1 sin 4146

B

rr r

rr r

r r

r

= ° − ° = °

° = ⇒ ° + ° = ⇒+

° = − °⇒°° = − ° ⇒ = ⇒

− °≈

40. 52 , 19B x= ° =

( )

90 52 38

cos 5219

19cos 52 cos 52

19cos 52 cos 52 1 cos 52

19cos5230

1 cos52

A

r

rr r

r r r

r

= ° − ° = °

° = ⇒+

° + ° = ⇒° = − ° = − ° ⇒

°= ≈− °

41. 62, 11y r= =

111 11tan tan 10

62 62

rA A

y− ⎛ ⎞= = ⇒ = ≈ °⎜ ⎟⎝ ⎠

( ) ( )2 22 2 2 2

2 2 2 2

11 11 62

11 11 62 11 62 11 52

x r r y x

x x

+ = + ⇒ + = + ⇒

+ = + ⇒ = + − ≈

42. 39, 20y r= =

120 20tan tan 27

39 39

rA A

y− ⎛ ⎞= = ⇒ = ≈ °⎜ ⎟⎝ ⎠

( ) ( )2 22 2 2 2

2 2

2 2

20 20 39

20 20 39

20 39 20 24

x r r y x

x

x

+ = + ⇒ + = + ⇒

+ = + ⇒

= + − ≈


43..

tan 73.8 3.4420 258ft

75 75° = ⇒ ≈ ⇒ ≈a a

a

44.

sin 40 0.6428 19ft

30 30° = ⇒ ≈ ⇒ ≈a a

a



45.

tan 24 35 0.4575 43ft93 93

a ac° = ⇒ ≈ ⇒ ≈′

46.

195 195

sin 28 4 0.4705 414ftcc c

° = ⇒ ≈ ⇒ ≈′

47.

145 45tan tan 37.8

58 58θ θ − ⎛ ⎞= ⇒ = ≈ °⎜ ⎟⎝ ⎠

48.

30 30

tan 21 78 fttan 21

xx

° = ⇒ = ≈°

49.

3.5 3.5

tan 8 25 fttan 8

xx

° = ⇒ = ≈°

50.

sin 52 27sin 52 21 ft27

xx° = ⇒ = ° ≈

51.

35 35

tan 20.5 94 fttan 20.5

xx

° = ⇒ = ≈°

52.

12.5 2.5tan tan 22.0

6.2 6.2θ θ − ⎛ ⎞= ⇒ = ≈ °⎜ ⎟⎝ ⎠


53.

cot cot , and

cot cot .

cot cotcot cot

(cot cot )cot cot

d xh d x

hx

h xh

h d hh h d

dh d h

α α

β β

α βα β

α βα β

+= ⇒ − =

= ⇒ =

− = ⇒− = ⇒

− = ⇒ =−

54.

cot cot , and cot

cot .cot cot cot cot

(cot cot )cot cot

α α β

βα β α β

α βα β

−= ⇒ = = ⇒

− == − ⇒ + = ⇒

+ = ⇒ =+

x d xh x

h hd h xh d h h h d

dh d h



55.

Using the result from exercise 54, we have 9

4.43cot 40 cot 50

= ≈° + °

h . Then

1(4.43)(9) 20

2= ≈ABCA square units.

56.

ABC is an isosceles triangle, so

52.5∠ = ∠ = °m B m C .

sin 52.5 7.910

° = ⇒ ≈ADAD .

cos 52.5 6.0876 12.210

° = ⇒ ≈ ⇒ ≈CDCD BC

1(12.2)(7.9) 48

2≈ ≈ABCA square units.

57.

One way to solve this is to use the results of exercise 54 to find x.

cot cot

10 3cot 30 cot 60

10 3 cot 30 10 3 cot 60

310 3 3 10 3 30 10 40

3

dh

x

x

α β=

+

=° + °

= ° + °⎛ ⎞

= ⋅ + = + =⎜ ⎟⎝ ⎠

58.

cos cos

cos

r dr r d

rd r r

θ θθ

−= ⇒ = − ⇒

= −

59.

Draw OA and perpendicular OH. Now find the measure of each central angle. From geometry, we know that the sum of the central angles in a

circle is 360°. Then, 360

,7

BOA°∠ = and

1 360.

2 14AOH BOH AOB

°∠ = ∠ = ∠ = Then,

360sin sin

14360 360

sin 2 2 sin .14 14

AH AHAOH

OA r

AH r AH AB r

°∠ = ⇒ = ⇒

° °= ⇒ = =

60.

( )( )2

sin 2 sin2

cos 2 cos2

1 12 sin 2 cos

2 2

2 sin cos

ABC

BC BCBC r

AB rAC AC

AC rAB r

A BC AC r r

r

θ θ

θ θ

θ θ

θ θ

= = ⇒ =

= = ⇒ =

= ⋅ =

=

△



61.

2 28 15 17AB = + =

a. ,ABC ACD∼△ △ so ,Bα = and

15cos cos .

17Bα = =

b. ,ABC CBD∼△ △ so ,Aβ = and

15tan tan .

8Aβ = =


62. Find the length of the third side using the Pythagorean theorem. Then, use the appropriate trigonometric function (based on the given sides) to find the measure of one of the acute angles. Finally, subtract the measure of this angle from 90° to find the measure of the other acute angle.

63. A right triangle formed given the measures of the two acute angles is not unique. Therefore, the length of the sides cannot be determined. Only the ratios of the sides can be determined.

2.3 Additional Applications of Right Triangles

2.3 Practice Problems

1.

1opposite 7 7tan tan 60

adjacent 4 4θ θ − ⎛ ⎞= = ⇒ = ≈ °⎜ ⎟⎝ ⎠

The bearing is N 60° W.

2.

From example 2, we know that 64 .PQR∠ = °

We are seeking PQ. 56 56

sin 64 62 misin 64

PRPQ

PQ PQ° = = ⇒ = ≈

°

3.

From example 3, we know that 401.x ≈ Then

tan 31.2 401tan 31.2 243 ft401

hh° = ⇒ = ° ≈

4.

From example 4, we know that 15.5 ft.h ≈ 15.5

sin 21.2

15.543 ft

sin 21.2

h

AD AD

AD

° = = ⇒

= ≈°


Section 2.3 Additional Applications of Right Triangles 65

5.

( )

sin 87.14.7

sin 87.1 4.7 sin 87.1

4.7 sin 87.1 sin 87.1

4.7 sin 87.1 1 sin 87.1

4.7 sin 87.13665 mi

1 sin 87.1

r

rr r

r r

r

r

° =+

= ° + °° = − °° = − °

°= ≈− °

2.3 A Exercises: Basic Skills and Concepts

1. A bearing is the measure of a(n) acute angle from due north or due south.

2. The bearing that indicates the direction 35° northwest of a given location is written N 35° W.

3. If you walk at a steady rate for 15 minutes at a bearing of N 45° E and turn and walk 15 more minutes (at the same rate) at a bearing of N 45° W, your final position will be due north of your starting position.

4. Janet rides her bike at a steady rate for ten minutes at a bearing of S 30° E, then turns and rides her bike due north for ten more minutes (at the same rate). To return to her starting point from the second location, she should ride at a bearing of S 75° W.

5. True

6. False. If you turn 120° clockwise from due north, then the bearing is S 60° E.

Use the figure for exercises 7–10.

7. A: N 10° E; B: N 75° E

8. C: S 70° E; D: S 18° E

9. F: S 35° W; G: S 65° W

10. H: N 83° W; I: N 17° W


11. N 45° E 12. N 45° W

13. due south


14. 12 1tan tan 27

4 2θ θ −− ⎛ ⎞= ⇒ = ≈ °⎜ ⎟⎝ ⎠−

The bearing is S 27° W.



15. 12 2tan tan 34

3 3θ θ − ⎛ ⎞= ⇒ = ≈ °⎜ ⎟⎝ ⎠−

The bearing is S 34° E.

16. ( )15tan tan 5 79

1θ θ −−= ⇒ = ≈ °

The bearing is N 79° W.


17.

18 8tan tan 34

12 12θ θ − ⎛ ⎞= ⇒ = ≈ °⎜ ⎟⎝ ⎠

The bearing is S 34° W.

18.

1200 200tan tan 59

120 120θ θ − ⎛ ⎞= ⇒ = ≈ °⎜ ⎟⎝ ⎠

The bearing isN 59° E.

19.

cos 40.7 180cos 40.7 136.5

180

sin 40.7 180sin 40.7 117.4180

yy

xx

° = ⇒ = ° ≈

° = ⇒ = ° ≈

The plane flew about 136.5 miles north and 117.4 miles east.

20.

cos 37.4 165cos37.4 131.1

165

sin 37.4 165sin 37.4 100.2165

yy

xx

° = ⇒ = ° ≈

° = ⇒ = ° ≈

The plane flew about 131.1 miles south and 100.2 miles west.

21.

The ship sailed 2 12 24 mi⋅ = from A to B at

N 49.7° E, and then 3 15 45 mi⋅ = from B to C at N 40.3° W. Since the axes are parallel,

49.7 .EAB DBA∠ = ∠ = ° 180

49.7 40.3 90DBA ABC CBF

ABC ABC∠ + ∠ + ∠ = °⇒

° + ∠ + °⇒ ∠ = °

Since ABC△ is a right triangle, we can use the Pythagorean theorem to find AC.

2 224 45 51AC = + = The ship is 51 miles from its starting point.

22.

The hiker walked 1.5 4 6 mi⋅ = at S 13.4° W

from A to B, and then 2 5 10 mi⋅ = from B to C at N 76.6° W. Since the axes are parallel,

13.4 .FBA BAD∠ = ∠ = °





76.6 13.4 90CBF FBA∠ +∠ = ° + ° = ° Since ABC△ is a right triangle, we can use the Pythagorean theorem to find AC.

2 26 10 11.7AC = + ≈ The hiker is about 11.7 miles from his starting point.

23.

The first oil rig is located at O, the second oil

rig is located at B, and the boat is located at A. Since the axes are parallel,

20.6 .OAB CBA∠ = ∠ = ° 3 3

tan 20.6 7.98tan 20.6

OAOA

° = ⇒ = ≈°

The boat is about 7.98 miles from the first rig.

24.

The first radar station is located at O, the

second radar station is located at B, and the plane is located at A. We are seeking AB.

14 14cos 61.7 29.5

cos 61.7AB

AB° = ⇒ = ≈

°

The plane is about 29.5 miles from the northernmost station.

25.

After one hour, the first ship is located at A, 14 miles from O, and the second ship is located at B, 20 miles from O. We are seeking .ABC∠ 67.3 22.7 180 90AOB AOB° + ∠ + ° = °⇒ ∠ = ° In right triangle AOB,

114 14tan tan 35

20 20OBA OBA − ⎛ ⎞∠ = ⇒ ∠ = ≈ °⎜ ⎟⎝ ⎠

Since the axes are parallel, 67.3 ,OBC∠ = ° so 67.3 35 32.3ABC OBC OBA∠ = ∠ − ∠ = ° − ° = °

The bearing from the second ship to the first ship after one hour is S 32.3° E.

26.

After two hours, the first plane is located at A,

600 miles from O, and the second plane is located at B, 900 miles from O. We are seeking

.CBA∠ 53.2 36.8 180 90AOB AOB° + ∠ + ° = °⇒ ∠ = ° In right triangle AOB,

1

600tan

9002

tan 33.7 .3

ABO

ABO −

∠ = ⇒

⎛ ⎞∠ = ≈ °⎜ ⎟⎝ ⎠

Since the N-S axes are parallel, 36.8 .CBO∠ = °

36.8 33.7 3.1CBA CBO ABO∠ = ∠ − ∠

= ° − ° ≈ °

The bearing from the second plane to the first plane after two hours is N 3.1° W.

In exercises 27–48, we don’t evaluate intermediate results in order to minimize rounding errors.

27.

The bear is originally located at C, and then

walks to D. We are seeking DC. In triangle

ABD, 35 35

tan 9.9 .tan 9.9

BDBD

° = ⇒ =°





In triangle ABC, 35 35

tan 6.5 .tan 6.5

BCBC

° = ⇒ =°

35 35106.7 ft

tan 6.5 tan 9.9DC BC BD= − = − ≈

° °

The bear walked 106.7 feet.

28.

The boat is originally located at A and then

moves to B. We are seeking AB = h.

In triangle ABD, tan 25100

h

BC° = ⇒

+

tan 25 100 tan 25 .h BC= ° + ° In triangle ABC,

tan 41 50 tan 41 50 .h

h BCBC

° = ⇒ = °′ ′ Now

equate the expressions for h and solve for BC.

( )

tan 25 100 tan 25 tan 41 50100 tan 25 tan 41 50 tan 25100 tan 25 tan 41 50 tan 25

100 tan 25.

tan 41 50 tan 25

BC BCBC BCBC

BC

° + ° = ° ⇒′° = ° − °⇒′° = ° − ° ⇒′

°=° − °′

Substitute this expression into tan 41 50h BC= ° ′ and solve for h.

100 tan 25tan 41 50 97m

tan 41 50 tan 25h

°⎛ ⎞= ° ≈′⎜ ⎟⎝ ⎠° − °′

The Statue of Liberty is about 97 m tall.

29.

The balloon is at A, the top of the statue at E,

and the bottom of the statue at D. The ground is CD. BCDE can be shown to be a rectangle. We are seeking ED. In triangle ABE,

30tan11.3

AB BC

BE BE

−° = = ⇒

30 30.

tan11.3 tan11.3

BC EDBE

− −= =° °

In triangle ACD,

30 30tan 26.6 .

tan 26.6

ACBE

DC BE° = = ⇒ =

°

Equate the expressions for BE, and solve for

ED. 30 30

tan11.3 tan 26.6

ED− = ⇒° °

30 tan11.330

tan 26.630 tan11.3

30 18tan 26.6

ED

ED ED

°− = ⇒°

°− = ⇒ ≈°

The statue is about 18 feet tall.

30.

The backdrop is represented by AC and the

singer is represented by BC. CD is the stage. We are seeking BC. In triangle ACD,

8.5 8.5tan13.6 .

tan13.6

ACCD

CD CD° = = ⇒ =

°

In triangle BCD,

tan 8.7 tan 8.7 .BC

BC CDCD

° = ⇒ = °

Substituting, we have 8.5

tan 8.7 5.4tan13.6

BC = ⋅ ° ≈°

The singer is about 5.4 feet tall.

31.

The building is represented by BC, the statue by

AB, and the ground by DC. We are seeking BC. In triangle ADC,

12.5tan 37.5

AC BC

DC DC

+° = = ⇒

12.5.

tan 37.5

BCDC

+=°

In triangle BDC,

tan 33.6 .tan 33.6

BC BCDC

DC° = ⇒ =

°

Equate the two expressions for DC and solve

for BC. 12.5

tan 37.5 tan 33.6

BC BC+ = ⇒° °

12.5 tan 33.6 tan 33.6 tan 37.5BC BC° + ° = °⇒





( )12.5 tan 33.6 tan 37.5 tan 33.612.5 tan 33.6 tan 37.5 tan 33.6

12.5 tan 33.680.7

tan 37.5 tan 33.6

BC BCBC

BC

° = ° − °⇒° = ° − ° ⇒

°= ≈° − °

The building is about 80.7 m tall.

32.

The cliff is represented by BC, the tree by AB,

and the distance from the cliff to the tourist’s boat by DC. We are seeking AB. In triangle

ADC, 40

tan 62.1AC AB

DC DC

+° = = ⇒

40.

tan 62.1

ABDC

+=°

In triangle BDC,

40 40tan 57.9 .

tan 57.9

BCDC

DC CD° = = ⇒ =

°

Equate the two expressions for DC and solve

for AB. 40 40

tan 62.1 tan 57.9

AB+ = ⇒° °

40 tan 57.9 tan 57.9 40 tan 62.1tan 57.9 40 tan 62.1 40 tan 57.9

40 tan 62.1 40 tan 57.97.4

tan 57.9

ABAB

AB

° + ° = °⇒° = ° − °⇒

° − °= ≈°

The tree is about 7.4 m tall.

33.

The bottom of the painting is located at D. We

are seeking BD. Note that BD = BC + CD. CD = 5.5 − 4 = 1.5 ft. Find BC as follows:

tan 50 12 tan 50 14.312

BCBC° = ⇒ = ° ≈ . The

height of the painting is 14.3 + 1.5 = 15.8 feet.

34. See diagram in next column. The height of the building AE = 1250 − BC. In BDE , we have

1250tan 67.7

DE° = ⇒ 1250

tan 67.7DE AC= =

°.

Then in BCA , we have

tan 55.8 .tan 55.8

BC BCAC

AC° = ⇒ =

°

Equate the two expressions for AC and solve

for BC. 1250

tan 67.7 tan 55.8

BC= ⇒° °

1250 tan 55.8.

tan 67.7BC

°=°

1250 tan 55.81250 495.6

tan 67.7AE

°≈ − ≈°

.

The building is about 495.6 feet tall.

35.

We are seeking DB.

3960 3960cos89 3 238,844

cos89 3AB

AB° = ⇒ = ≈′

° ′

238,844 3960 234,884DB AB AD= − = − = The distance from the moon to the surface of the earth is about 234,884 miles.

36.

We are seeking DS. r = 3960.

3960 3960sin10.1 .

sin10.1AS

AS° = ⇒ =

°

39603960 18,621

sin10.1DS AS AD= − = − ≈

°

The satellite is about 18,621 miles from the surface of the earth.



37.

We are seeking CD. In triangle ABC,

tan 72.6 25 tan 72.6 .25

ACAC° = ⇒ = °

In triangle ACD,

tan 21.825 tan 72.6

25 tan 72.6 tan 21.8 32

CD CD

ACCD

° = = ⇒°

= ° ° ≈

The silo is about 32 feet tall.

38.

Juan is located at J and Kathy at K. We are

seeking AC. In triangle JKC,

tan 60.2 50 tan 60.2 .50

JCJC° = ⇒ = °

In triangle ACJ,

tan 50.150 tan 60.2

50 tan 60.2 tan 50.1 104.4

AC AC

JCAC

° = = ⇒°

= ° ° ≈

The tower is about 104.4 feet tall.

39.

From geometry, we know that the radius OB is

perpendicular to AB since a radius drawn to the point of tangency is perpendicular to the

tangent. Also, from geometry, we know that 12

11.75 .OAB CAB∠ = ∠ = °

tan11.75 5 tan11.75 .5

OBOB° = ⇒ = ° The

diameter = 2 2 5 tan11.75 2 feet.OB = ⋅ ° ≈

40.

From geometry, we know that

12

14 .BAO BAC∠ = ∠ = ° In triangle BAO,

tan14 12 tan14 .12

OBOB° = ⇒ = ° Since OB is a

radius, the diameter = 2 2 12 tan14 6 ft.OB = ⋅ ° ≈

41.

In order to find the area, we need to find BC.

Using the Pythagorean theorem, we have 2 2 2

2 2 2

2 2

210.3 87.72

210.3 87.72

AC AB BC

BC

BC

= + ⇒= + ⇒

= −

The area is 2 2210.3 87.72 87.72 16,766BC AC⋅ = − ⋅ ≈

square meters.

42.

Since triangle ABC is equilateral, we know that

each angle measures 60°. In triangle ADC, 1.2 1.2

sin 60 1.4sin 60

ACAC

° = ⇒ = ≈°

Each side is about 1.4 feet long.



43.

From geometry, we know that the diagonals of

a rhombus are perpendicular to each other and bisect each other. They also bisect the angles of the rhombus. Thus, angle DAE measures 24° and ED = 0.8m. In triangle DAE,

0.8sin sin 24

0.82.0

sin 24

EDDAE

AD AD

AD

∠ = ⇒ ° = ⇒

= ≈°

The length of each side is about 2.0 m.

44.

The area of the parallelogram is 20h.

sin 49.3 16sin 49.3 .16

hh° = ⇒ = ° Thus, the

area = 20 16sin 49.3 242.6 sq m.⋅ ° ≈

45.

The height of the plane = BC + CD. We don’t

have information about any of the sides in TCD . We know that TC = AB and

8.1∠ = ∠ = °m CTB m ABT . Then 150

tan 8.1 1053.955° = ⇒ ≈ABAB

. Now find

CD: tan 67.4 2531.9641053.955

° ≈ ⇒ ≈CDCD .

The height of the plane is 150 2532 2682 feet.BC CD+ ≈ + ≈

46.

The height of the building AE = 600 − BC.

tan 75.24° = BC

AC and

600tan 60.05

−° = BC

AC.

( )

( )

600So, and

tan 75.24 tan 60.05600

tan 75.24 tan 60.05tan 60.05 tan 75.24 600

tan 60.05 600 tan 75.24 tan 75.24

tan 60.05 tan 75.24 600 tan 75.24

600 tan 75.24

tan 60.05 tan 75

BC BCAC AC

BC BC

BC BC

BC BC

BC

BC

−= = ⇒° °−= ⇒

° °° = ° − ⇒° = ° − °⇒° + ° = °⇒

°=° +

411.7329.24

≈°

.

Then 600 411.7329 188.27≈ − ≈AE meters. The building is about 188.27 m tall.

For exercises 47 and 48, we use the result of exercise

2.2.53, .cot cot

dh

α β=

−

47.

The plane is located at D, and the tracking

stations are located at A and B. We are seeking h. 500

5402 mcot 36.9 cot 38.9

h = ≈° − °

48.

The building is located at B, the plane’s first

position at E, and the plane’s second position at D. We are seeking DE. In triangle BED,

12,000 40,337 ftcot10.4 cot 25.6

= ≈° − °

d





Since the plane traveled 40,337 feet in two minutes, its speed is 40,337 2 20,168.5= feet

per minute.


49.

Draw altitude OH. Then 12

,AH AB= and the

area of triangle AOB = 12

.AB OH AH OH⋅ = ⋅

Since angle AOB = 120°, angle AOH =

60°. sin sin 6010

AH AHAOH

OA∠ = ° = = ⇒

10sin 60 .

cos cos 6010

10cos 60 .

AHOH OH

AOHOA

OH

= °

∠ = ° = = ⇒

= °

The area of the triangle = 10sin 60 10cos 60° ⋅ °

23 1100 25 3 cm .

2 2

⎛ ⎞ ⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

50.

The area of the triangle is 12

.bh

sin sin .h

h aa

θ θ= ⇒ =

1 1Area sin sin .

2 2b a abθ θ= ⋅ =

51.

Because OB = OC = OA, BCOα = ∠ and .ACOβ = ∠

.ACB ACO BCO α β∠ = ∠ + ∠ = +

In triangle ABC,

( )180

180 2 2 18090 .

A ACB Bα α β β α βα β

∠ + ∠ +∠ = °⇒+ + + = °⇒ + = °⇒+ = °

Thus, angle ACB is a right angle.

52.

In right triangle ADE,

tan 31 20 tan 31 .20

DEDE° = ⇒ = °

In right triangle BDE, 20 tan 31

tan 39DE

BE BE

°° = = ⇒

20 tan 31.

tan 39BE

°=°

In right triangle CEB, 20 tan 31

tan 39tan 35BE

CE DE x

°°° = =

+

20 tan 31

tan 3920 tan 31

20 tan 31 tan 35 tan 35

20 tan 31

tan 39tan 35

20 tan 3120 tan 31 tan 35

tan 3920 tan 31

20 tan 31 tan 35tan 39 9.2

tan 35

xx

x

x

°°= ⇒° +

° ° + °°= ⇒

°°

°= − ° °⇒°° − ° °

°= ≈°



53.

a. 180

180180

540 (1)(2)

(3)(4)

BAC ACB BCAD ACD ADCEAD EDA EBAC ACB B CAD ACD

ADC EAD EDAE

BAC CAD DAE BAEACB ACD BCDADC EDA CDE

∠ +∠ + ∠ = °∠ + ∠ + ∠ = °∠ + ∠ + ∠ = °∠ + ∠ + ∠ + ∠ + ∠

+∠ +∠ + ∠+∠ = °

∠ + ∠ + ∠ = ∠∠ + ∠ = ∠∠ +∠ = ∠

Substitute (2), (3), and (4) into (1): 540

5405 540 108

5

BAE E CDE BCD BBAE E CDE BCD B

B B

∠ + ∠ + ∠ + ∠ + ∠ = °∠ = ∠ = ∠ = ∠ = ∠

°∠ = °⇒ ∠ = = °

b. BAC ACB∠ = ∠ since triangle ABC is isosceles. 2 180B BAC∠ + ∠ = °⇒ 108 2 180 2 72

36BAC BAC

BAC° + ∠ = °⇒ ∠ = °⇒

∠ = °

Similarly, 36 .DAE∠ = °

36 36 108 36BAC CAD DAE BAE

CAD CAD∠ +∠ + ∠ = ∠ ⇒° + ∠ + ° = °⇒ ∠ = °

For parts (c) and (d), draw altitudes AM and BN. The altitudes bisect CD and AC, respectively.

c. In triangle ABN,

12 1

cos cos36 .2

b bNAB

a a∠ = ° = = ⋅

d. In triangle ACM,

12

1cos cos 36 cos18

2

1.

2

CAM

a a

b b

⎛ ⎞∠ = ⋅ = °⎜ ⎟⎝ ⎠

= = ⋅

54.

a. In triangle DBC,

tan tan 45DC

DBCBC

∠ = ° = ⇒

11 1.BC

BC= ⇒ =

Using the Pythagorean theorem, 2 2 2 21 1 2.BD BC DC= + = + =

b. We are given AB = DB, so

2 1.AC AB BC= + = + Use the Pythagorean theorem in triangle ADC to find AD.

( )( )

22 2 21 2 1

1 2 2 2 1 4 2 2

AD DC AC= + = + +

= + + + = +

c. Angle DBA = 180° – 45° = 135°. Triangle ABD is isosceles, so angle A = angle BDA.

1802 135 180 2 45

22.5

A BDA DBAA A

A

∠ + ∠ + ∠ = °⇒∠ + ° = °⇒ ∠ = °⇒∠ = °

d. In triangle ADC,

1 2cos cos 22.5 .

4 2 2

ACA

AD

+∠ = ⇒ ° =+

55.

a. In triangle DBC,

tan tan 30DC

DBCBC

∠ = ° = ⇒

3 1 33.

3 3BC

BC= ⇒ = =





Using the Pythagorean theorem,

( )22 2 21 3 1 3

4 2.

BD DC BC= + = + = +

= =

b. We are given AB = DB, so

2 3.AC AB BC= + = + Use the Pytha-gorean theorem in triangle ADC to find AD.

( )( )

( )

22 2 21 2 3

1 4 4 3 3 8 4 3

4 2 3 2 2 3.

AD DC AC= + = + +

= + + + = +

= + = +

In order to show that 2 2 3 6 2,+ = +

start by squaring 2 2 3 :+

( ) ( )22 2 3 4 2 3 8 4 3.+ = + = +

Now square 6 2 :+

( )26 2 6 2 12 2 8 2 4 3

8 4 3.

+ = + + = + ⋅

= +

( ) ( )2 22 2 3 8 4 3 6 2+ = + = + ⇒

2 2 3 6 2+ = + (since both numbers are positive.)

c. Angle DBA = 180° – 30° = 150°. Triangle ABD is isosceles, so angle A = angle BDA.

1802 150 180 2 30

15

A BDA DBAA A

A

∠ + ∠ +∠ = °⇒∠ + ° = °⇒ ∠ = °⇒∠ = °

d. In triangle ADC,

2 3cos cos15 .

6 2

ACA

AD

+∠ = ⇒ ° =+


56. Use this figure to help solve parts (a)–(d).

a. N 35° E

b. 180° – 150° = 30° S 30° E

c. 240° – 180° = 60° S 60° W

d. 360° – 310° = 50° N 50° W

Chapter 2 Review Exercises

For exercises 1 and 2, recall the following.

opposite hypotenusesin csc

hypotenuse oppositeadjacent hypotenuse

cos sechypotenuse adjacentopposite adjacent

tan cotadjacent opposite

θ θ

θ θ

θ θ

= =

= =

= =

1.

3 3 13 13sin csc

13 3132 2 13 13

cos sec13 213

3 2tan cot

2 3

θ θ

θ θ

θ θ

= = =

= = =

= =

2.

1 17sin csc 17

17174 4 17 17

cos sec17 417

1tan cot 4

4

θ θ

θ θ

θ θ

= = =

= = =

= =

3. ( )cos 43 sin 90 43 sin 47° = ° − ° = °

4. ( )sec32 csc 90 32 csc58° = ° − ° = °

5. ( )sin 24 cos 24 90θ θ° = − ° ⇒ = °


Chapter 2 Review Exercises 75

6. ( )tan cot 17θ θ= + °

The definition of a cofunction gives

( )tan cot 90 .θ θ= ° − Equate the two

expressions for tan ,θ then solve for .θ

( ) ( )cot 17 cot 9090 17 2 73 36.5

θ θθ θ θ θ+ ° = ° − ⇒

° − = + °⇒ = °⇒ = °

7. ( ) ( )sec 4 csc 36θ θ+ ° = + °


( ) ( )( )( )

sec 4 csc 90 4

csc 86 .

θ θθ

+ ° = ° − + °= ° −

Equate the two expressions for ( )sec 4 ,θ + °

then solve for .θ

( ) ( )csc 36 csc 8636 86 2 50 25θ θ

θ θ θ θ+ ° = ° − ⇒

+ ° = ° − ⇒ = °⇒ = °

8. ( )cos 2 sin 21θ θ= + °


( )cos 2 sin 90 2 .θ θ= ° −

Equate the two expressions for cos 2 ,θ then solve for .θ

( ) ( )sin 21 sin 90 221 90 2 3 69 23θ θ

θ θ θ θ+ ° = ° − ⇒

+ ° = ° − ⇒ = °⇒ = °

For exercises 9–10, first find the length of the missing side using the Pythagorean theorem, then find the trigonometric functions of A using

sin csc

cos sec

tan cot

a cA A

c ab c

A Ac ba b

A Ab a

= =

= =

= =

9. ( )22 2 2 2 24 4 3 64

8

c a b c

c

= + ⇒ = + = ⇒=

4 1 8sin csc 2

8 2 44 3 3 8 2 3

cos sec8 2 34 34 3 4 3

tan cot 33 44 3

A A

A A

A A

= = = =

= = = =

= = = =

10. 2 2 2 2 2 2 26 5 11

11

c a b b b

b

= + ⇒ = + ⇒ = ⇒=

5 6sin csc

6 511 6 6 11

cos sec6 11115 5 11 11

tan cot11 511

A A

A A

A A

= =

= = =

= = =

For exercises 11–20, we use a calculator to determine the value of each expression.

11. sin 46.19 0.7216° ≈

12. tan 68.42 2.5283° ≈

13. 1

sec 4.37 1.0029cos 4.37

° = ≈°

14. cos 78.22 0.2042° ≈

15. 1

sin14.2 0.2453csc14.2

= ° ≈°

16. ( )cot 90 46.24 tan 46.24 1.0442° − ° = ° ≈

17. sin 0.9659258263 75.0θ θ≈ ⇒ ≈ °

18. cos 0.0348994937 88.0θ θ≈ ⇒ ≈ °

19. tan 1.07236871 47.0θ θ≈ ⇒ ≈ °

20. sec 1.3242692451

cos 52.01.324269245

θ

θ θ

≈ ⇒

≈ ⇒ ≈ °

21.

90 36 54B = ° − ° = °

sin sin 3618

18sin 36 11in.

cos cos 3618

18cos36 15 in.

a aA

ca

b bA

cb

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈



22.

90 46.47 43.53B = ° − ° = °

28.61sin sin 46.47

28.6139.46

sin 46.4728.61

tan tan 46.47

28.6127.18

tan 46.47

aA

c c

c

aA

b b

b

= ⇒ ° = ⇒

= ≈°

= ⇒ ° = ⇒

= ≈°

Use the triangle below to help you identify the opposite and adjacent legs in exercises 23–28.

23. A = 21.8°, c =1.92 cm 90 21.8 68.2

sin sin 21.81.92

1.92sin 21.8 0.713 cm

cos cos 21.81.92

1.92cos 21.8 1.78 cm

B

a aA

ca

b bA

cb

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈

24. A = 71.6°, c = 6.89 cm 90 71.6 18.4

sin sin 71.66.89

6.89sin 71.6 6.54 cm

cos cos 71.66.89

6.89cos 71.6 2.17 cm

B

a aA

ca

b bA

cb

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈

25. A = 61.42°, b = 15.2 ft 90 61.42 28.58

tan tan 61.5215.2

15.2 tan 61.42 27.9 ft

B

a aA

ba

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

15.2cos cos 61.42

15.231.8 ft

cos 61.42

bA

c c

c

= ⇒ ° = ⇒

= ≈°

26. A = 42.37°, b = 26.9 ft 90 42.37 47.63

tan tan 42.3726.9

26.9 tan 42.37 24.5 ft

26.9cos cos 42.37

26.936.4 ft

cos 42.37

B

a aA

ba

bA

c c

c

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ≈°

27. a = 7.28 m, b = 20.42 m 2 2 2 2

1

7.28 20.42 21.7 m7.28

tan tan20.42

7.28tan 19.6

20.4290 19.6 70.4

c a ba

A Ab

A

B

−

= + = + ≈

= ⇒ = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °

28. a = 13.9 m, b = 14.2 m 2 2 2 2

1

13.9 14.2 19.9 m13.9

tan tan14.2

13.9tan 44.4

14.290 44.4 45.6

c a ba

A Ab

A

B

−

= + = + ≈

= ⇒ = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠≈ ° − ° ≈ °


29. 35 , 51 , 19A BDC y= ° ∠ = ° =


BDCx

∠ = ⇒

tan 51 tan 51 .h

h xx

° = ⇒ = °


Ay x

∠ = ⇒+

( )tan 35 19 tan 35 .19

hh x

x° = ⇒ = + °

+





Now equate the two expressions for h, and solve for x.

( )tan 51 19 tan 35x x° = + °⇒

( )

tan 51 tan 35 19 tan 35

tan 51 tan 35 19 tan 35

tan 51 tan 35 19 tan 35

19 tan 3525

tan 51 tan 35tan 51 25 tan 51 31

x x

x x

x

x

h x

° = ° + °⇒° − ° = °⇒° − ° = °⇒

°= ≈° − °

= ° = ° ≈

30. 15 , 27 , 14A DBC y= ° ∠ = ° =


DBCh

∠ = ⇒

tan 27 .tan 27

x xh

h° = ⇒ =

°


Ay x

∠ = ⇒+

( )tan15 14 tan15 .14

hh x

x° = ⇒ = + °

+

Now equate the two expressions for h, and solve for x.

( )14 tan15tan 27

xx= + °⇒

°

( )

tan15 14 tan15tan 27

tan15 tan 27 14 tan15 tan 27

tan15 tan 27 14 tan15 tan 27

1 tan15 tan 27 14 tan15 tan 27

14 tan15 tan 272.2

1 tan15 tan 272.2

4.3tan 27 tan 27

xx

x x

x x

x

x

xh

= ° + °⇒°

= ° ° + ° ° ⇒− ° ° = ° ° ⇒− ° ° = ° ° ⇒

° °= ≈− ° °

= = ≈° °

31. 129 , 51 , 30A BDC c= ° ∠ = ° =


sinh

Ac

= ⇒

sin 29 30sin 29 15.30

hh° = ⇒ = ° ≈


BDCx

∠ = ⇒

15 15tan 51 12.

tan 51x

x° = ⇒ = ≈

°

In right triangle ABC, tan .h

Ay x

=+

We use the

intermediate values to avoid additional rounding

errors.

( )

( )

30sin 29tan 29

15

tan 5115

tan 29 tan 29 30sin 29tan 51

15tan 29 30sin 29 tan 29

tan 51

1530sin 29 tan 29

tan 5114.

tan 29

y

y

y

y

°° = ⇒+

°⎛ ⎞° + ° = °⇒⎜ ⎟⎝ ⎠°

⎛ ⎞° = ° − ° ⇒⎜ ⎟⎝ ⎠°⎛ ⎞° − ° ⎜ ⎟⎝ ⎠°= ≈

°

32. 232 , 38 , 20A BDC c= ° ∠ = ° =


sinh

BDCc

∠ = ⇒

sin 38 20sin 38 12.20

hh° = ⇒ = ° ≈


cosx

BDCc

∠ = ⇒

cos 38 20cos 38 16.20

xx° = ⇒ = ° ≈

In right triangle ABC, tan .h

Ay x

=+

We use the

intermediate values to avoid additional rounding errors.

20sin 38tan 32

20cos 38tan 32 20 tan 32 cos 38 20sin 38

tan 32 20sin 38 20 tan 32 cos38

20sin 38 20 tan 32 cos 383.9

tan 32

y

y

y

y

°° = ⇒+ °

° + ° ° = °⇒° = ° − ° ° ⇒

° − ° °= ≈°


33. 31 , 9A r= ° =

90 31 59

9sin 31 sin 31 9sin 31 9

99 9sin 31

sin 31 9 9sin 31 8sin 31

B

xx

x x

= ° − ° = °

° = ⇒ ° + ° = ⇒+

− °° = − °⇒ = ≈°



34. 48 , 18B x= ° =

( )

90 48 42

sin 4218

sin 42 18sin 4218sin 42 sin 42

18sin 42 1 sin 42

18sin 4236

1 sin 42

A

r

rr r

r r

r

r

= ° − ° = °

° = ⇒+

° + ° = ⇒° = − °⇒° = − ° ⇒

°= ≈− °

35. 51, 15y r= =

115 15tan tan 16

51 51A A − ⎛ ⎞= ⇒ = ≈ °⎜ ⎟⎝ ⎠

15sin sin16

1515sin16 sin16 15

sin16 15 15sin1615 15sin16

39sin16

rA

r x xx

x

x

= ⇒ ° = ⇒+ +° + ° = ⇒° = − °⇒− °= ≈

°

36. (5, 0) is due east of the origin.

37. (3, 3) is N 45° E of the origin.

38. (0, –4) is due south of the origin.

39. (–1, –2) lies in quadrant III. First find the

reference angle. 2

tan 2 63.41

α α−= = ⇒ ≈ °−

90 63.4 26.6θ = ° − ° = °

(–1, –2) is S 26.6° W of the origin.

40. (–5, 5) is N 45° W of the origin.

41.

We are seeking AB. Since the triangle is a 30°-

60°-90° triangle, we know that the length of the hypotenuse is twice the length of the shorter leg. Thus, AB = 2(40) = 80 ft.

42.

We are seeking AC. Since the triangle is an

isosceles right triangle, we know that the length of the hypotenuse equals the length of a leg

times 2. Thus, 2 2AC= ⇒ 2

2 1.4.2

AC = = ≈ The tip of the pendulum

is about 1.4 feet from the vertical line through its pivot.

43.

The spotlight is located at A and the beam of

light hits the wall at B. We are seeking AC. 7 7

tan 40 8.3tan 40

ACAC

° = ⇒ = ≈°

The light should be placed about 8.3 feet from the wall.

44.

The building is represented by BC and the ladder by AB. We are seeking BC.

sin 7524

24sin 75 23.2

BC

BC

° = ⇒

= ° ≈

The ladder reaches about 23.2 feet up the building.

45.

The engineer is located at A, and the monument

is represented by BC. We are seeking BC.

tan 34.1 115 tan 34.1 77.9115

BCBC° = ⇒ = ° ≈

The monument is about 77.9 feet tall.



46.

From geometry, we know that the diagonals of

a rhombus bisect each other and the vertex angles. We are given that the length of one diagonal is 1.8 m and that the top angle measures 50°. Therefore, angle ABD = 25° and AD = 0.9 m. The sides of a rhombus are equal, so we are seeking AB.

0.9 0.9sin 25 2

sin 25AB

AB° = ⇒ = ≈

°

The sides of the rhombus are about 2 m long.

47.

First, use the Pythagorean theorem to find the

width of the rectangle. 2 2 2 2285.6 218.5

183.9BC AC AB= − = −

≈

The area of the rectangle equals (AB)(BC). ( )( ) ( )( )218.5 183.9 40,185 sq mAB BC = ≈

48.

90 37.4 52.6

sin sin 52.6160

160sin 52.6 127

cos cos 52.6160

160cos 52.6 97

BOA

AB ABBOA

OBAB

AO AOBOA

OBAB

∠ = ° − ° = °

∠ = ⇒ ° = ⇒

= ° ≈

∠ = ⇒ ° = ⇒

= ° ≈

The plane is about 127 miles north and 97 miles west from the airport.

49.

74 in.

tan 35 106xx

° = ⇒ ≈

The shadow is about 106 inches or 8 feet 10 inches long.

50.

tan12 1885

xx° = ⇒ ≈

The tree is about 18 feet tall.

51.

850

tan 40 1013xx

° = ⇒ ≈

The rescue ship is about 1013 ft from the crippled boat.

52.

After one hour, the first ship is located at A, 12

miles from O, and the second ship is located at B, 18 miles from O. We are seeking .CBA∠ 21.3 68.7 180 90AOB AOB° + ∠ + ° = °⇒ ∠ = ° In right triangle AOB,

112 2tan tan 33.7 .

18 3ABO ABO − ⎛ ⎞∠ = ⇒ ∠ = ≈ °⎜ ⎟⎝ ⎠

Since the N-S axes are parallel, 68.7 .CBO∠ = °

68.7 33.7 35CBA CBO ABO∠ = ∠ − ∠

= ° − ° ≈ °

The bearing from the second ship to the first ship after two hours is N 35° W.



53.

The bottom of the bust is located at D. We are

seeking BD. Note that BD = BC + CD. CD = 5 ft 2 in. − 4 ft = 1 ft 2 in.. Find BC as follows:

tan 30 6 tan 30 3.56

BCBC° = ⇒ = ° ≈ . The

height of the bust is 3 ft 6 in. + 1 ft 2 in. = 4 ft 8 in.

54.

The plane started at O, flew to A at N 22° W,

then turned 90° and flew to B. We are seeking .θ In right triangle OAB,

2.5tan 2.5 68.2 .

1BOA BOA∠ = = ⇒ ∠ ≈ °

68.2 22 46.2θ = ° − ° = ° The bearing from the airport to the plane is N 46.2° E.

55.

The obelisk is represented by MN. Kaitlyn is

originally located at A, and then moves to B. In triangle ABN,

tan 82.9 30 tan 82.9 .30

ANAN° = ⇒ = °

In triangle AMN,

tan 430 tan 82.9

30 tan 82.9 tan 4 16.8

MN MN

ANMN

° = = ⇒°

= ° ° ≈

The obelisk is about 16.8 feet tall.

Chapter 2 Test

1. 9 3 12 4

sin , cos ,15 5 15 59 3

tan12 4

A A

A

= = = =

= =

2. Since sin 18° = cos (90° – 18°) = cos 72°, 1

cos 72 .1 5

° =+

3. ( )cos 20 sin 2 20θ° = − °

The definition of a cofunction gives ( )cos 20 sin 90 20 sin 70 .° = ° − ° = °

Equate the two expressions for cos 20 ,° then solve for .θ 2 20 70 2 90 45θ θ θ− ° = °⇒ = °⇒ = °

4.

First, use the Pythagorean theorem to find c.

2 25 12 13c = + = 5 12 5

sin , cos , tan13 13 12

A A A= = =

5. Using a calculator, 1

sec 76.531 4.2933.cos 76.531

° = ≈°

6.

21.0θ = °

7. A = 28.7°, right angle C, c = 12.9 in. 90 28.7 61.3

sin sin 28.712.9

12.9sin 28.7 6.19 in.

cos cos 28.712.9

12.9cos 28.7 11.3 in.

B

a aA

ca

b bA

cb

= ° − ° = °

= ⇒ ° = ⇒

= ° ≈

= ⇒ ° = ⇒

= ° ≈


Chapter 2 Test 81

8. a = 17.68 cm, b = 22.19 cm, right angle C. 2 217.68 22.19 28.37 cmc = + ≈

1

17.68tan tan

22.1917.68

tan 38.5522.19

90 38.55 51.45

aA A

b

A

B

−

= ⇒ = ⇒

⎛ ⎞= ≈ °⎜ ⎟⎝ ⎠= ° − ° = °

9.

In right triangle BCD,

tan 54 tan 54 .h

h xx

° = ⇒ = °

In right triangle ABC,

tan 32 20 tan 32 tan 32 .20

hh x

x° = ⇒ = ° + °

+

Equate the two expressions for h, and solve for x.

( )

tan 54 20 tan 32 tan 32tan 54 tan 32 20 tan 32tan 54 tan 32 20 tan 32

20 tan 3217 ft

tan 54 tan 32

x xx xx

x

° = ° + °⇒° − ° = °⇒° − ° = °⇒

°= ≈° − °

10.

A = 32°, r = 10

90 32 5810

sin sin 3210

10sin 32 sin 32 10

sin 32 10 10sin 3210 10sin 32

8.9sin 32

B

rA

r x xx

x

x

= ° − ° = °

= ⇒ ° = ⇒+ +° + ° = ⇒° = − °⇒− °= ≈

°

11.

The boat is located at A and the dock at B. We

are seeking AB.

4

sin sin14

417

sin14

BCA

AB AB

AB

= ⇒ ° = ⇒

= ≈°

The rope is about 17 ft long.

12.

100 100

cos10 101.5 ftcos10

xx

° = ⇒ = ≈°

13.

In right triangle ABO,

90 30 60 ,AOB∠ = ° − ° = ° and 30 .OAB∠ = ° In a 30°-60°-90° triangle, the leg opposite the 30° angle is half the hypotenuse, and the leg opposite the 60° angle equals the shorter leg

times 3. Thus, OB = 70, and

70 3 121.AB = ≈ The plane flew about 70 miles west and about 121 miles north.

14.

After one hour, the first ship is located at A, 12

miles from O, and the second ship is located at B, 16 miles from O. We are seeking .CBA∠ 15 75 180 90 .AOB AOB° + ∠ + ° = °⇒ ∠ = ° In right triangle AOB,

112 3tan tan 36.9 .

16 4ABO ABO − ⎛ ⎞∠ = ⇒ ∠ = ≈ °⎜ ⎟⎝ ⎠





Since the N-S axes are parallel, 75 .CBO∠ = °

75 36.9 38.1CBA CBO ABO∠ = ∠ − ∠

= ° − ° ≈ °

The bearing from the second ship to the first ship after two hours is N 38.1° W.

15.

The mural is represented by BD, and the

camera is located at A. In right triangle ACD,

tan tan 2110

10 tan 21 3.8.

CD CDCAD

ACCD

∠ = ⇒ ° = ⇒

= ° ≈

In right triangle ABC,

tan tan 3810

10 tan 38 7.8.

BC BCCAB

ACCD

∠ = ⇒ ° = ⇒

= ° ≈

BD = BC + CD = 3.8 + 7.8 = 11.6 The mural is about 11.6 feet tall.


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