Precalculus mathematics for calculus international metric edition 7th edition stewart solutions manu

Precalculus Mathematics for Calculus International Metric Edition 7th Edition Stewart Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/precalculus-mathematics-for-calculus-international-metric-edition-7th-edition-st ewart-solutions-manual/

6

TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH

6.1

ANGLE MEASURE

1. (a) The radian measure of an angle is the length of the arc that subtends the angle in a circle of radius 1. . (b) To convert degrees to radians we multiply by 180

(c) To convert radians to degrees we multiply by 180 .

2. (a) If a central angle is drawn in a circle of radius r, the length of the arc subtended by is s r . (b) The area of the sector with central angle is A 12 r 2 .

3. (a) The angular speed of the point is $

. t

s (b) The linear speed of the point is ) . t (c) The linear speed ) and the angular speed ome are related by the equation ) r$.

4. No, if the common angular speed is $, Object A has linear speed 2$, while Object B has angular speed 5$. Object B has greater linear speed. rad rad s 0 262 rad 5. 15i 15i 180 i 12

rad rad s 0 628 rad 6. 36i 36i 180 i 5

rad 3 rad s 0 942 rad 7. 54i 54i 180 i 10

rad 5 rad s 1 309 rad 8. 75i 75i 180 i 12

rad rad s 0 785 rad 9. 45i 45i 180 i 4

rad rad s 0 524 rad 10. 30i 30i 180 i 6

rad 5 rad s 1 745 rad 11. 100i 100i 180 i 9

rad 10 rad s 3 491 rad 12. 200i 200i 180 i 9

rad 50 rad s 17 453 rad 13. 1000i 1000i 180 i 9

rad 20 rad s 62 832 rad 14. 3600i 3600i 180 i

rad 7 rad s 1 222 rad 15. 70i 70i 180 i 18

rad 5 rad s 2 618 rad 16. 150i 150i 180 i 6

i

i 18. 34 34 180 135

i

i

i 20. 32 32 180 270

i 17. 53 53 180 300

i

i 19. 56 56 180 150 i

i

540 i 21. 3 3 180 s 171 9 i

i

216 i 23. 1 2 1 2 180 68 8 i

180 18i 25. 10

10 i

2 180 24i 27. 215

15

i

i

360 i 22. 2 2 180 s 114 6 i

i

612 i 24. 3 4 3 4 180 s 194 8 i

5 180 50i 26. 518

18 i

13 180 i 28. 13 12 12 195

29. 50i is coterminal with 50i 360i 410i , 50i 720i 770i , 50i 360i 310i , and 50i 720i 670i . (Other answers are possible.)

30. 135i is coterminal with 135i 360i 495i , 135i 720i 855i , 135i 360i 225i , and 135i 720i 585i .

(Other answers are possible.)

31. 34 is coterminal with 34 2 114 , 34 4 194 , 34 2 54 , and 34 4 134 . (Other answers are possible.)

11 13 32. 116 is coterminal with 116 2 236 , 116 4 356 , 116 2 6 , and 6 4 6 . (Other answers

are possible.)

477

478

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

7

15

9

17 33.

4 is coterminal with 4 2 4 , 4 4 4 , 4 2 4 , and 4 4 4 . (Other answers are possible.)

34. 45i is coterminal with 45i 360i 315i , 45i 720i 675i , 45i 360i 405i , and 45i 720i 765i .

(Other answers are possible.)

35. Since 430i 70i 360i , the angles are coterminal.

36. Since 330i 30i 360i , the angles are coterminal.

37. Since 176 56 126 2 ; the angles are coterminal.

38. Since 323 113 213 7 is not a multiple of 2 , the angles are not coterminal.

39. Since 875i 155i 720i 2 360i , the angles are coterminal.

40. Since 340i 50i 290i is not a multiple of 360i , the angles are not coterminal.

41. Since 400i 360i 40i , the angles 400i and 40i are coterminal.

42. Since 375i 360i 15i , the angles 375i and 15i are coterminal.

43. Since 780i 2 360i 60i , the angles 780i and 60i are coterminal.

44. Since 100i 260i 360i is a multiple of 360i , the angles 100i and 260i are coterminal.

45. Since 800i 3 360i 280i , the angles 800i and 280i are coterminal.

46. Since 1270i 190i 1080i 3 360i is a multiple of 360i , the angles 1270i and 190i are coterminal.

47. Since 196 2 76 , the angles 196 and 76 are coterminal.

5

48. Since 53 2 3 , the angles 3 and 3 are coterminal.

49. Since 25 12 2 , the angles 25 and are coterminal.

50. Since 10 2 s 3 717, the angles 10 and 10 2 are coterminal.

17

51. Since 174 2 2 4 , the angles 4 and 4 are coterminal.

52. Since 512 32 24 12 2 , the angles 512 and 32 are coterminal.

53. Using the formula s r, s 56 9 152 .

r 35 7 s 5 s 12 2.

5 54. Using the formula s r, the length of the arc is s 140i 180i 9 9 i s 10 180 55.

2 rad 2 s 114 6i r 5

8 s 56. Solving for r, we have r , so the radius of the circle is r 4.

2 57. Solving for s, we have s r , so the length of the arc is 5 3 15 cm.

8 58. Solving for s, we have s r 12 40i

s 8 38 m.

180i 3 i 14 14 180 s rad

s 89 1i .

59. Solving for , we have r 9 9

s 5 5 180i 60. Solving for , we have rad s 95 5i . r 3 3

15 18 s

s 5 73 m 61. r

5 6

s 20 72 22 92 cm 62. r i

i

50 180

32 4 128 s 44 68 63. (a) A 12 r 2 12 82 80i 180 i 9 9

(b) A 12 r 2 12 102 0 5 25 U U 2A 2 12 , so the radius is s 5 86.

64. (a) A 12 r 2 " r

0 7

SECTION 6.1 Angle Measure

U (b) r

2A

V

479

2 12 s 3 03 150i 180 i

r s

2 65. 23 rad and r 10 m, so A 12 r 2 12 10 2 23 100 3 s 104 7 m . r s

1 2 1 2 29 29 s 5 1 m 2.

29 66. 145i 145i 36 and r 2 m , so A 2 r 2 2 36 18 180i

7 1 7 2 rad. Thus, A 12 r 2 % 70

r "

67. A 70 m2 and 140i 140i 180i 9 2 9 T U 6 5 9

s 7 6 m. r 2 70 7

T U , so A 1 r 2 % 20 5 r 2 " r 2 20 12 4 6 s 5 5 m.

68. A 20 m2 and 512 2 12 5

69. r 80 km and A 1600 km2, so A 12 r 2 % 1600 12 80 2 % 12 rad.

U 600 2 2 . Thus, the area of the sector is 70. The area of the circle is r 600 m , so r

900 1 600 3 s 286 5 m2 .

A 12 r 2 2

71. Referring to the figure, we have AC 3 1 4, BC 1 2 3, and AB 2 3 5. Since AB 2 AC 2 BC 2 , then by the Pythagorean Theorem, the

triangle is a right triangle. Therefore, 2 and 2 A 12 r 2 12 12 2 4 m .

#

72. The triangle is equilateral, so 1 3 rad. To find 2 , we use the formula 2 i 1 2 3 1 s 0 047 rad, or approximately 2 7 .

$

%

s 1

1 rad. Thus, r 1

73. Between 1:00 P. M . and 1:00 P. M ., the minute hand traverses three-quarters of a complete revolution, or 34 2 32 rad, while the hour hand moves three-quarters of the way from 12 to 1, which is itself one-twelfth of a revolution. So the hour r s 1 2 rad. hand traverses 34 12 8 74. Between 1:00 P. M . and 6:45 P. M ., the minute hand traverses five complete revolutions plus three-quarters of a revolution; that is, 5 2 34 2 232 rad, while the hour hand moves through five-twelfths of a revolution, plus three-quarters of r s 5 2 3 1 2 23 rad. the way from 6 to 7; that is, 12 4 12 24 75. The circumference of each wheel is d 70 cm. If the wheels revolve 10,000 times, the distance traveled is 1m 1 km 10,000 70 cm

s 21 99 k m.

100 cm 1000 m 76. Since the diameter is 75 cm, we have r 37. 5 cm . In one revolution, the arc length (distance traveled) is s r 2 37. 5 75 c m . The total distance traveled is 1 k m 1000 m/k m 100 cm/m 100,000 cm 100 ,00 0 cm 1 rev s 424 41 rev. Therefore the car wheel will make approximately 424 revolutions.

75 cm rad rad.

77. We find the measure of the angle in degrees and then convert to radians. 40 5i 25 5i 15i and 15 180 i 12

6340 528 s 1659 808 a nd so the distance between the two cities is Then using the formula s r , we have s 12 roughly 1660 k m. rad rad. Then using the formula s r , the length of the arc is 78. 35i 30i 5i 5i 180 i 36

s 36 6340 176 s 553 27. So the distance between the two cities is roughly 553 km.

480

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

1 of its orbit which is 2 rad. Then s r 2 149 ,000,000 s 2 ,564 ,916 7 , so the 79. In one day, the earth travels 365 365 365

distance traveled is approximately 2 5 6 million kilometers.

80. Since the sun is so far away, we can assume that the rays of the sun are parallel when striking the earth. Thus, the angle 800 s 180 8 00 formed at the center of the earth is also 7 2i . So r 7 2 s 6 368 k m, and the circumference

7 2i 180 i 2 180 8 00 40 ,000 k m.

7 2 r s 1 i 1 rad rad. Then s r 6 34 0 s 1 844 , and so a 81. The central angle is 1 minute 60 60 180i 10,800 10,800 is c 2 r

nautical mile is approximately 1 844 k m. b c s 19,791 m2 .

82. The area is A 12 r 2 12 902 280i 180 i 83. The area is equal to the area of the large sector (with radius 85 cm ) minus the area of the small sector (with radius 35 cm ) r sb c s 7069 cm 2 .

Thus, A 12 r12 12 r22 12 852 35 2 135i 180 i 84. The area available to the cow is shown in the diagram. Its area is the sum of four quarter-circles: s r A 14 302 152 122 92

m

m

337.5

m m m

s 1060 m2 m m m

45 2 rad 90 rad/min.

1 min 45 2 40 (b) The linear speed is ) 3600 cm /min s 11,309 7 cm /min 1 1000 2 rad 2000 rad/min.

86. (a) The angular speed is $ 1 min

85. (a) The angular speed is $

(b) The linear speed is ) 87. )

1000 2 115 00 5 m /s s 15 7 m /s.

60

8 2 1 16

s 3.351 m/s.

15 15

600 2 128 00 m 1 km 60 min 20 16 k m/h s 6 3 3 k m/h.

1 min 1 hr 1000 m 1 day 1 2 6340

s 1664 36 k m/h.

89. 23 h 56 min 4 s 23 9344 hr. So the linear speed is 1 day 23 9344 hr

88. )

linear speed 8 0 k m/h 1 h 1000 m

1111 rad/min . radius 1. 2 m 60 min 1 k m angular speed 1111 rad/min (b) The rate of revolution is

s 177 rev/min.

2 2 100 2 0 20 m 2 91. )

s 2 09 m/s.

60 s 3 40 2 10 linear speed of pedal

92. (a) The angular speed is $ 160 rad/min. radius of wheel sprocket 5 (b) The linear speed of the bicycle is ) angular speed radius 160 rad/min 32 .5 cm 5 200 cm /min s 9 8 k m/h.

90. (a) The radius is 1. 2 m , so the angular speed is $

SECTION 6.2 Trigonometry of Right Triangles

481

93. (a) The circumference of the opening is the length of the arc subtended by the angle on the flat piece of paper, that is, C s r 6 53 10 s 31 4 cm.

10 C

5 cm.

2 2 T (c) By the Pythagorean Theorem, h 2 62 52 11, so h 11 s 3 3 cm.

T (d) The volume of a cone is V 13 r 2 h. In this case V 13 52 11 s 86 8 cm3 . (b) Solving for r, we find r

94. (a) With an arbitrary angle , the circumference of the opening is V S 3 C 9 2 2 2

, h 6 r 36 2 , and C 6 , r 2

V S 9 2 9 2 9 V 13 r 2 h 2 36 2 2 2 4 2 2 .

3

(b)

100 50 0 0

2

4

6

(c) The volume seems to be maximized for s 5 13 rad or about 293i .

95. Answers will vary, although of course everybody prefers radians.

6.2 1. (a)

TRIGONOMETRY OF RIGHT TRIANGLES (b) sin

opposite adjacent

adjacent opposite opposite , cos , and tan . hypotenuse hypotenuse adjacent

(c) The trigonometric ratios do not depend on the size of the triangle because all right triangles with angle are similar.

hypotenuse

n

2. The reciprocal identities state that csc

1 1 1 , sec , and cot . sin cos tan

3. sin 45 , cos 35 , tan 43 , csc 54 , sec 53 , cot 34 7 , cos 24 , tan 7 , csc 25 , sec 25 , cot 24 4. sin 25 7 7 25 24 24 S T 9 2 5. The remaining side is obtained by the Pythagorean Theorem: 41 402 81 9. Then sin 40 41 , cos 41 , 41 41 9 tan 40 9 , csc 40 , sec 9 , cot 40

6. The hypotenuse is obtained by the Pythagorean Theorem:

S T 8 82 152 289 17. Then sin 15 17 , cos 17 ,

17 17 8 tan 15 8 , csc 15 , sec 8 , cot 15

7. The remaining side is obtained by the Pythagorean Theorem: T

T

S T 32 22 13. Then sin T2

13

T

T

2 1313 ,

cos T3 3 1313 , tan 23 , csc 213 , sec 313 , cot 32 13 T S T 8. The remaining side is obtained by the Pythagorean Theorem: 82 72 15. Then sin 78 , cos 815 , T

T

T

tan T7 7 1515 , csc 87 , sec T8 8 1515 , cot 715 15 15 S T 2 2 9. c 5 3 34 T

(a) sin cos T3 3 3434 34 S T 2 2 10. b 7 4 33

(b) tan cot 35

T

(c) sec csc 534

482

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

(a) sin cos 47

(b) tan cot T4

(c) sec csc T7

33

33

11. (a) sin 22i s 0 37461

(b) cot 23i s 2 35585

12. (a) cos 37i s 0 79864

(b) csc 48i s 1 34563

13. (a) sec 13i s 1 02630

(b) tan 51i s 1 23490

14. (a) csc 10i s 5 75877 x , we have x 25 sin 30i 25 12 25 15. Since sin 30i 2.

25 T 12 12 12 , we have x 1 12 2.

16. Since sin 45i i x sin 45 T

(b) sin 46i s 0 71934

2

T T x 17. Since sin 60i , we have x 13 sin 60i 13 23 132 3 .

13 T 4 4 4 1 4 3.

18. Since tan 30i , we have x x tan 30i T 3

12 12 19. Since tan 36i s 16 51658. , we have x x tan 36i 25 25 20. Since sin 53i , we have x s 31 30339. x sin 53i x y 21. cos % x 28 cos , and sin % y 28 sin .

28 28 4 4 x tan % x 4 tan , and cos % y 4 sec .

22.

4 y cos S T 24. cos 12 23. tan 56 . Then the third side is x 52 62 61.

13 . The third side is S T T T y 132 122 25 5. The other five ratios are The other five ratios are sin 5 6161 , cos 6 6161 , T

T

csc 561 , sec 661 , and cot 65 .

5 , tan 5 , csc 13 , sec 13 , and sin 13 12 5 12

cot 12 5.

n

n

25. cot 1. Then the third side is r

S T 12 12 2.

T

The other five ratios are sin T1 22 , 2 T T cos T1 22 , tan 1, csc 2, and 2 T sec 2.

26. tan

S T 3. The third side is r 12 3 2. The other T

five ratios are sin 23 , cos 12 , csc T2 , 3 sec 2, and cot T1 .

3

n

n

SECTION 6.2 Trigonometry of Right Triangles

27. csc 11 6 . The third side is x

S S T T 112 62 85. The 28. cot 53 . The third side is x 52 32 34. The T

T

T

6 , cos 85 , other five ratios are sin 11 11

other five ratios are sin 3 3434 , cos 5 3434 ,

tan 6 8585 , sec 118585 , and cot 685 .

tan 35 , csc 334 and sec 534 .

T

T

T

T

T

n

n

T T 1 3 1 3 29. sin

cos 6 6 2 2 2

30. sin 30i csc 30i sin 30i

483

1 1 sin 30i

T

T

31. sin 30i cos 60i sin 60i cos 30i 12 12 23 23 14 34 1 r T s2 r s2 32. sin 60i 2 cos 60i 2 23 12 34 14 1 33. cos 30i 2 sin 30i 2

r T s2 3 2

r s2 12 34 14 12

s2 K rT rT r rT sL2 s2 s T b 3 T1 T1 1 sin cos c2 1 1 1 3 2 3 1 T 34. sin cos

3

1

3

1

3 4 4 3 8 8 2T s 2 2r r2 T 2s 2 18 4 2 3 14 2 3 T s2 rT r srT s b 2 1 2 3 2 c2 1 1 2 1

sin

35. cos 4 6 2 2 2 4 2 2 4 2 rT T s2 r s2 T T T c b 3 3 2 36. sin 3 tan 12 2 94 2 6 csc 4 2 3 2

37. This is an isosceles right triangle, so the other leg has length 16 tan 45i 16, the hypotenuse has length T 16 16 2 s 22 63, and the other angle is 90i 45i 45i . i sin 45 100 s 103 52, and the other angle is 38. The other leg has length 100 tan 75i s 26 79, the hypotenuse has length sin 75i i i i 90 75 15 .

35 s 56 85, and the other angle is 39. The other leg has length 35 tan 52i s 44 80, the hypotenuse has length cos 52i i i i 90 52 38 .

40. The adjacent leg has length 1000 cos 68i s 374 61, the opposite leg has length 1000 sin 68i s 927 18, and the other angle is 90i 68i 22i .

41. The adjacent leg has length 33 5 cos 8 s 30 95, the opposite leg has length 33 5 sin 8 s 12 82, and the other angle is 3 .

2

8

8

42. The opposite leg has length 72 3 tan 6 s 41 74, the hypotenuse has length .

2

6

72 3 s 83 48, and the other angle is cos 6

3

43. The adjacent leg has length

106 106 s 145 90, the hypotenuse has length s 180 34, and the other angle is tan sin 5 5

3 .

2

5

10

44. The adjacent leg has length 425 cos 38 s 162 64, the opposite leg has length 425 sin 38 s 392 65, and the other angle is 3 .

2

8

8

484

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

1 s 0 45 cos s 2 s 0 89, tan 1 , csc s 2 24, sec s 2 24 s 1 12, cot s 2 00.

45. sin s 2 24 2 24 2 2

sin 40i s 0 64, cos 40i s 0 77, tan 40i s

46.

0 64 s 0 83, csc 40i s 1 56, 0 77

sec 40i s 1 31, cot 40i s 1 20.

c

47. x

100 100

s 230 9 tan 60i tan 30i

48. Let d be the length of the base of the 60i triangle. Then tan 60i %d x

85 85 85 %d s 49 075, and so tan 30i d tan 60i d x

85 85 %x d s 98 1. tan 30i tan 30i

50 h 50 h %h s 57 735 % sin 65i % x s 63 7 h sin 60i x sin 65i 5 5 h %h % 10, and so tan 60i 50. Let h be the hypotenuse of the top triangle. Then sin 30i h sin 30i x h 10 x

s 5 8. tan 60i tan 60i y x , so x y sin 10 sin tan .

51.

From the diagram, sin and tan y 10 49. Let h be the length of the shared side. Then sin 60i

n

y n

Z

Y

n

a b 1 d % a sin , tan % b tan , cos % c sec , cos % d cos 1 1 c 1 h i % h 16 00 tan 11i s 311 m .

53. Let h be the height, in meters , of the Empire State Building. Then tan 11 1600 10,500 54. (a) Let r be the distance, in meters , between the plane and the Gateway Arch. Therefore, sin 22i % r 10,500 r s 28 ,029 m . sin 22i (b) Let x be the distance, also in meters , between a point on the ground directly below the plane and the Gateway Arch. Then 10,500 10,500 %x s 25,988 m. tan 22i x tan 22i h 55. (a) Let h be the distance, in kilometers, that the beam has diverged. Then tan 0 5i % 384,000 h 384 ,000 tan 0 5i s 3351 k m.

52. sin

(b) Since the deflection is about 3351 k m whereas the radius of the moon is about 16 00 k m, the beam will not strike the moon.

60 60 %x s 1 41 m .

56. Let x be the distance, in meters , of the ship from the base of the lighthouse. Then tan 23i x tan 23i h % h 6 sin 72i s 5.7 m .

57. Let h represent the height, in meters , that the ladder reaches on the building. Then sin 72i 6 h % h 180 sin 65i s 163 m .

58. Let h be the height, in meters, of the communication tower. Then sin 65i 180 h 59. Let h be the height, in meters , of the kite above the ground. Then sin 50i % h 135 sin 50i s 103 m.

135

SECTION 6.2 Trigonometry of Right Triangles

485

Let h 1 be the height of the flagpole above elevation and let h 2 be the height below,

60.

I| c Y c

Il

h as shown in the figure. So tan 18i 1 % h 1 x tan 18i . Similarly, x h 2 x tan 14i . Since the flagpole is 18 meters tall, we have h 1 h 2 18 , so x tan 18i tan 14i 18 % x

18 s 3 1 4 m . tan 18i tan 14i

61. Let h 1 be the height of the window in meters and h 2 be the height from the window to the top of the tower. Then tan 25i

h1 98

h2 % h 2 98 tan 39 i s 79 m . Therefore, the height of the window 98 is approximately 46 m and the height of the tower is approximately 46 79 12 5 m .

% h 1 98 tan 25i s 46 m . Also, tan 39 i

62.

Let d1 be the distance, in meters , between a point directly below the plane and one

c

c

car, and d2 be the distance, in meters , between the same point and the other car. Then

c

c

E|

tan 52i

El

1545 1545 1545 % d1 s 1207 09, and tan 35i % d1 tan 52i d2

1545 s 2206 49 m. So the distance between the two cars is about d1 d2 s 1207 0 9 2206 49 s 3414 m . tan 35i 63. Let d1 be the distance, in meters , between a point directly below the plane and one car, and d2 be the distance, in meters , between d d the same point and the other car. Then tan 52i 1 % d1 1 54 5 tan 52 i s 1977 5 m. Also, tan 38 i 2 % 1545 1545 d2 1 54 5 tan 38 i s 1207 09 m . So in this case, the distance between the two cars is about 77 0 m . d2

64. Let x be the distance, in meters , between a point directly below the balloon and the first kilometer post. Let h be the height, h h in meters , of the balloon. Then tan 22i and tan 20i . So h x tan 22i x 1000 tan 20i % x x 1000 1000 tan 20i s 9087 m . Therefore h s 9087 tan 22 i s 3671 m s 3 7 k m. x tan 22i tan 20i 65. Let x be the horizontal distance, in meters , between a point on the ground directly below the top of the mountain and h the point on the plain closest to the mountain. Let h be the height, in meters , of the mountain. Then tan 35i x h 1000 tan 32i i i i and tan 32 . So h x tan 35 x 1000 tan 32 % x s 8294 2. Thus x 1000 tan 35i tan 32i h s 8294 2 tan 35i s 5808 m .

Since the angle of elevation from the observer is 45i , the distance from the

66.

observer is h, as shown in the figure. Thus, the length of the leg in the smaller right I

c I

I

h % 600 h tan 75i h % 600 h 600 tan 75i s 473 m.

600 tan 75i h 1 tan 75i % h 1 tan 75i

triangle is 600 h. Then tan 75i

67. Let d be the distance, in kilometers, from the earth to the sun.

Then sec 89 85i

d % 384,000

d 384 ,000 sec 89 85 i s 146 7 million kilometers. i 68. (a) s r % rs 9848 6340 s 1 553 3 rad s 89

(b) Let d represent the distance, in kilometers , from the center of the earth to the moon. Since cos d

6340 s cos

6340 s 363, 273 7. So the distance AC is 363, 273 7 6 34 0 s 356,900 km . cos 89 i

6340 , we have d

486

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

69. Let r represent the radius, in kilometers, of the earth. Then sin 60 276i r r960 % r 9 60 sin 60 276i r % i

60 sin 60 276 s 6 33 6 158 . So the earth’s radius is about 6 34 0 k m.

960 sin 60 276i r 1 sin 60 276i % r 91 sin 60 276i

,000 70. Let d represent the distance, in kilo meters , from the earth to Alpha Centauri. Since sin 0 000211i 149,000 , we have d 149, 000,000 13 d sin 0 000211i s 40 ,46 5,283 ,160 ,456 . So the distance from the earth to Alpha Centauri is about 4 0 5 10 km.

d d, so d sin 46 3i s 0 723 AU. 1 72. If two triangles are similar, then their corresponding angles are equal and their corresponding sides are proportional. That n n n n n n is, if triangle ABC is similar to triangle A) B ) C ) then AB r n A) B ) n, AC r n A) C ) n, and BC r n B ) C ) n. Thus when we express any trigonometric ratio of these lengths as a fraction, the factor r cancels out.

71. Let d be the distance, in AU, between Venus and the sun. Then sin 46 3i

6.3

TRIGONOMETRIC FUNCTIONS OF ANGLES

1. If the angle is in standard position and P x y is a point on the terminal side of , and r is the distance from the origin to x y y P, then sin , cos , and tan . r r x 2. The sign of a trigonometric function of depends on the quadrant in which the terminal angle of lies. For example, if is in quadrant II, sin is positive. In quadrant III, cos is negative. In quadrant IV, sin is negative.

3. (a) If is in standard position, then the reference angle is the acute angle formed by the terminal side of and the x-axis.

So the reference angle for 100i is 80i and that for 190i is 10i . (b) If is any angle, the value of a trigonometric function of is the same, except possibly for sign, as the value of the trigonometric function of . So sin 100i sin 80i and sin 190i sin 10i .

4. The area A of a triangle with sides of lengths a and b and with included angle is given by the formula A 12 ab sin . So the area of the triangle with sides 4 and 7 and included angle 30i is 12 4 7 sin 30i 7.

5. (a) The reference angle for 120i is 180i 120i 60i .

6. (a) The reference angle for 175i is 180i 175i 5i .

(b) The reference angle for 200i is 200i 180i 20i .

(b) The reference angle for 310i is 360i 310i 50i .

(c) The reference angle for 285i is 360i 285i 75i .

(c) The reference angle for 730i is 730i 720i 10i .

7. (a) The reference angle for 225i is 225i 180i 45i . (b) The reference angle for 810i is 810i 720i 90i . (c) The reference angle for 105i is 180i 105i 75i . is 7 3 .

9. (a) The reference angle for 710 10 10

8. (a) The reference angle for 99i is 180i 99i 81i . (b) The reference angle for 199i is 199i 180i 19i . (c) The reference angle for 359i is 360i 359i 1i .

10. (a) The reference angle for 56 is 56 6.

(b) The reference angle for 98 is 98 8.

(b) The reference angle for 109 is 109 9.

(c) The reference angle for 103 is 103 3 3.

(c) The reference angle for 237 is 237 3 27 .

11. (a) The reference angle for 57 is 57 27 .

12. (a) The reference angle for 2 3 is 2 3 2 0 3 .

(b) The reference angle for 1 4 is 1 4 0 4 .

(b) The reference angle for 2 3 is 2 3 s 0 84.

(c) The reference angle for 1 4 is 1 4 because 1 4 2.

(c) The reference angle for 10 is 10 10 0.

T 13. cos 150i cos 30i 23

T

14. sin 240i sin 60i 23

SECTION 6.3 Trigonometric Functions of Angles T

15. tan 330i tan 30i 33

16. sin 30i sin 30i 12 T

T 17. cot 120i cot 60i 33

18. csc 300i csc 60i 2 3 3

T 3

19. csc 630i csc 90i sin190i 1

20. cot 210i cot 30i tan130i

21. cos 570i cos 30i 23

22. sec 120i sec 60i cos160i 2

T

T 23. tan 750i tan 30i T1 33 3

24. cos 660i cos 60i 12

25. sin 32 sin 2 1 s r T 27. tan 43 tan 3 3 s r b c 29. csc 56 csc 6 2 31. sec 173 sec 3

487

1 26. cos 43 cos 3 2 T s r 3 28. cos 116 cos

6 2 T

2 3 30. sec 76 sec 6 3

1 2 cos 3

32. csc 54 csc 4

T 1 2 sin 4 T

b c 1

33. cot 4 cot 4 tan 1 4

2 T1 34. cos 74 cos 4 2 2

35. tan 52 tan 2 which is undefined.

1 36. sin 116 sin 6 2

37. Since sin 0 and cos 0, is in quadrant III. 38. Since both tan and sin are negative, is in quadrant IV. sin 0 % sin 0 (since cos 0). Since sin 0 and cos 0, is 39. sec 0 " cos 0. Also tan 0 " cos in quadrant IV. 40. Since csc 0 " sin 0 and cos 0, is in quadrant II. S S sin 1 cos2 2

.

41. Since sin is negative in quadrant III, sin 1 cos and we have tan cos cos S 1 sin2 cos

because cos 0 in quadrant II. 42. cot sin sin S 43. cos2 sin2 1 % cos 1 sin2 because cos 0 in quadrant IV. 1 1 44. sec S because all trigonometric functions are positive in quadrant I. cos 1 sin2 S 45. sec2 1 tan2 % sec 1 tan2 because sec 0 in quadrant II. S 46. csc2 1 cot2 % csc 1 cot2 because csc 0 in quadrant III. S T 47. sin 45 . Then x 52 42 9 3, since is in quadrant IV. Thus, cos 35 , tan 43 , csc 54 , sec 53 , and cot 34 .

S 48. tan 43 , so r 42 32 5. Since is in quadrant III, x 3 and y 4, and so sin 45 , cos 35 , csc 54 , sec 53 , and cot 34 .

U T T T T r s2 7 95 7 7 49. cos 12 , so sin 1 12 1295 , tan 795 , csc 129595 , sec 12 , and cot

7 95 T T T S T 9 145 8 145 8 9 9 50. cot 9 , so r 82 92 145. Thus, sin T 145 , cos 145 , tan 8 , csc 145 9 , T and sec 145 8 .

51. csc 2. Then sin 12 and x T and cot 3.

145

T T T S T 22 12 3. So sin 12 , cos 23 , tan T1 33 , sec T2 2 3 3 , 3 3

488

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

T S T 52. cot 14 . Then tan 4 and r 42 12 17. So sin T4 , cos T1 , tan 4, csc 417 , and 17 17 T sec 17.

T T T S T T 7 7 5 , sec 7 , 53. cos 27 . Then y 72 22 45 3 5, and so sin 3 7 5 , tan 3 2 5 , csc T 15 2 3 5

T 2 2 5. and cot T 15 3 5

54. tan 4. Then r

T S T T 42 12 17, and so sin T4 , cos T1 , csc 417 , sec 17, and 17

17

cot 14 .

T T 3 2

55. If 3 , then sin 2 sin 3 2 and 2 sin 2 sin 3 3. r s b c 2 sin 2 3 and sin 2 sin 2 s 0 890.

56. If , then sin 3 3 4 9

57. The area is 12 7 9 sin 72i s 30 0.

58. The area is 12 10 22 sin 10i s 19 1.

T 59. The area is 12 10 2 sin 60i 25 3 s 43 3.

T

60. The area is 12 13 2 sin 60i 1694 3 s 73 2.

r s 1 32 1 154, or 61. Let the angle be . Then A 12 x y sin " 16 12 5 7 sin % sin 32 35 % sin 35 approximately 66 1i .

V

62. Let the lengths of the equal sides be x. Then A 12 x 2 sin % 24 12 x 2 sin 56 % x

48 sin 56

T T 96 4 6 s 9 8 cm.

4 . For the triangle defined by the two sides, 63. For the sector defined by the two sides, A1 12 r 2 12 22 120i 180 i 3 T T A2 12 ab sin 12 2 2 sin 120i 2 sin 60i 3. Thus the area of the region is A1 A2 43 3 s 2 46.

64. The area of the entire circle is r 2 122 144 , the area of the sector is 12 r 2 12 122 3 24 , T 1 1

2 and the area of the triangle is 2 ab sin 2 12 sin 3 36 3, so the area of the shaded region is T T A1 A2 A3 144 24 36 3 120 36 3 s 439 3.

65. (a) tan (b)

1000 m h 1600 tan m. , so h tan 1 . 6 km 1 km 1.6 kilometers

20i

60i

80i

85i

h

582

2771

9074

18,288

66. (a) Let the depth of the water be h. Then the cross-sectional area of the gutter is a trapezoid whose height is h 10 sin . The bases are 10 and 10 2 10 cos 10 20 cos . Thus, the area is b b2 10 10 20 cos A 1 h 10 sin 100 sin 100 sin cos .

2 2 (b)

(c)

200

130.0 129.5

100 0 0

1

129.0 1.00

1.05

1.10

From the graph, the largest area is achieved when s 1 047 rad s 60i .

489

SECTION 6.3 Trigonometric Functions of Angles

67. (a) From the figure in the text, we express depth and width in terms of .

(b)

depth width and cos , we have depth 20 sin 20 20 and width 20 cos . Thus, the cross-section area of the beam is A depth width 20 cos 20 sin 400 cos sin .

Since sin

200

0 0

(c) The beam with the largest cross-sectional area is the square beam, T T 10 2 by 10 2 (about 14 14 by 14 14).

1

68. Using depth 20 sin and width 20 cos , from Exercise 65, we have strength k width depth 2 . Thus S k 20 cos 20 sin 2 8000k cos sin2 .

69. (a) On Earth, the range is R

3.6 2 sin ) 02 sin 2 3 s 1 145 m and the height is

g 9.8

3.62 sin2 ) 2 sin2 6 9 0 1653 m .

H 0

2g 2 9.8 16 2 2 3.62 sin 3 s 7 015 m and H 3.6 sin 6 s 1.013 m (b) On the moon, R 2 1 6 1 6 U T 600 70. Substituting, t 4 15 s 15 5 s.

5 sin 30 i

71. (a) W 3 02 0 38 cot 0 65 csc

10

(b) From the graph, it appears that W has its minimum value at about 0 946 s 54 2i .

5 0 0

72. (a) We label the lengths L 1 and L 2 as shown in the

n

.l

3 2 figure. Since sin and cos , we L1 L2 have L 1 3 csc and L2 2 sec . Thus

n

L L 1 L 2 3 csc 2 sec . .|

(c) The minimum value of L is 7 023. (d) The minimum value of L is the shortest distance the pipe must pass through.

n

(b)

2

14 7 0 0.0

0.5

1.0

1.5

73. We have sin k sin , where 59 4i and k 1 33. Substituting, sin 59 4i 1 33 sin sin 59 4i s 0 6472. Using a calculator, we find that s sin 1 0 6472 s 40 3i , so " sin 1 33 4 2 s 4 40 3i 2 59 4i 42 4i .

74. sin 4i 0 0697564737 but sin 4 0 7568024953. Your partner has found sin 4i instead of sin 4 radians.

OP OP adj

O P . Since QS is tangent to the circle at R, -O R Q is a right triangle. Then 75. cos OR hyp 1 RQ OQ hyp opp R Q and sec O Q . Since / SO Q is a right angle -S O Q is a right

tan OR OR adj adj OS SR adj hyp O S and cot S R . Summarizing, we have

triangle and / O S R . Then csc OR OR opp opp sin P R , cos O P , tan R Q , sec O Q , csc O S , and cot S R .

490

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

r s 76. (a) sin2 cos2 1 % sin2 cos2

1 1 1 % tan2 1 sec2 cos2 cos2 r s 1 1 (b) sin2 cos2 1 % sin2 cos2 1 % 1 cot2 csc2 sin2 sin2

77. Let x represent the desired real number, in radians. We wish to find the smallest nonzero solution of the equation sin x sin 180x . (Since

Z

Y Z sin y

x is in radians, 180x is in degrees.) Now if a is in degrees with 0 n a n 180 and sin a b, then either a b or 180 a b. The graph shows that the first positive solution occurs when

Z sin Y

Y

@ 180x 180x n 180, so we have 180 x.

180 180 s 3 088. You can verify that the value of sin is the same whether Solving this equation, we get x 180 180 your calculator is set to radian mode or degree mode.

0

6.4

INVERSE TRIGONOMETRIC FUNCTIONS AND TRIANGLES

1. For a function to have an inverse, it must be one-to-one. To define the inverse sine function we restrict the domain of the d e sine function to the interval 2 2 . d e 2. (a) The function sin 1 has domain [ 1 1] and range 2 2 . (b) The function cos 1 has domain [ 1 1] and range [0 ]. b c (c) The function tan 1 has domain R and range 2 2 .

8 sin 1 4 6 cos 1 3 (b) cos 1 10 3. (a) sin 1 10 5 5 s r 5 , we let cos 1 5 and 4. To find sin cos 1 13 13

complete the right triangle as shown. We find that r s 5 12 . sin cos 1 13 13

(c) tan 1 86 tan 1 43

n

T 3 3

5. (a) sin 1 1 2

(b) cos 1 0 2

(c) tan 1

6. (a) sin 1 0 0 r T s 7. (a) sin 1 22 4

(b) cos 1 1 r T s (b) cos 1 22 34

(c) tan 1 0 0 (c) tan 1 1 4

r T s 8. (a) sin 1 23 3

r s (b) cos 1 12 23

r T s (c) tan 1 3 3

7. sin 1 0 30 0 30469

r s 10. cos 1 0 2 s 1 77215 11. cos 1 13 s 1 23096

12. sin 1 65 s 0 98511

13. tan 1 3 s 1 24905

14. tan 1 4 s 1 32582 15. cos 1 3 is undefined.

16. sin 1 2 is undefined.

6 3 , so sin 1 3 s 36 9i .

17. sin 10 5 5

7 , so tan 1 7 s 21 3i .

18. tan 18 18

9 , so tan 1 9 s 34 7i .

19. tan 13 13

3 1 3 s 25 4i .

20. sin 30 7 70 7 , so sin

21. sin 47 , so sin 1 47 s 34 8i .

22. cos 89 , so cos 1 89 s 27 3i .

SECTION 6.4 Inverse Trigonometric Functions and Triangles

491

d e 23. We use sin 1 to find one solution in the interval 90i 90i . sin 23 " sin 1 23 s 41 8i . Another solution with between 0i and 180i is obtained by taking the supplement of the angle: 180i 41 8i 138 2i . So the solutions of the equation with between 0i and 180i are approximately 41 8i and 138 2i .

d e 24. One solution is given by cos 1 43 s 41 4i . This is the only solution, because cos x is one-to-one on 0 180i . r s e d 25. One solution is given by cos 1 25 s 113 6i . This is the only solution, because cos x is one-to-one on 0 180i . e d 26. tan 1 20 s 87 1i , so the only solution in 0 180i is approximately 180 87 1 92 9i . d e 27. tan 1 5 s 78 7i . This is the only solution on 0 180i . r s d e 28. One solution is sin 1 45 s 53 1i . Another solution on 0 180i is approximately 180i 53 1i 126 9i . s r 29. To find cos sin 1 45 , first let sin 1 54 . Then is the number in the interval d e 2 2 whose sine is 45 . We draw a right triangle with as one of its acute

angles, with opposite side 4 and hypotenuse 5. The remaining leg of the triangle is

n

found by the Pythagorean Theorem to be 3. From the figure we get r s cos sin 1 54 sin 35 .

s s r r Another method: By the cancellation properties of inverse functions, sin sin 1 54 is exactly 45 . To find cos sin 1 45 , we

first write the cosine function in terms of the sine function. Let u sin 1 54 . Since 0 n u n 2 , cos u is positive, and since U T r s U r s2 T S 9 3 cos2 u sin2 u 1, we can write cos u 1 sin2 u 1 sin2 sin 1 54 1 45 1 16 25 25 5 . s r Therefore, cos sin 1 54 35 . s r 30. To find cos tan 1 54 , we draw a right triangle with angle , opposite side 4, and r s adjacent side 3. From the figure we see that cos tan 1 34 cos 35 .

n

s r 12 , we draw a right triangle with 31. To find sec sin 1 13 angle , opposite side 12, and hypotenuse 13. From the r s 12 sec 13 . figure we see that sec sin 1 13 5

n

s r 7 , we draw a right triangle with 32. To find csc cos 1 25 angle , opposite side 7, and hypotenuse 25. From the r s 7 csc 25 . figure we see that csc cos 1 25 24

n

492

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

s s r r 12 , we draw a right triangle with angle 34. To find cot sin 1 2 , we draw a right triangle with angle 33. To find tan sin 1 13 3 , opposite side 12, and hypotenuse 13. From the figure r s 12 tan 12 . we see that tan sin 1 13 5

n

, opposite side 2, and hypotenuse 3. From the figure we T s r see that cot sin 1 32 cot 25 .

n

r s 35. We want to find cos sin 1 x . Let sin 1 x, so sin x. We sketch a right triangle with an acute angle , opposite side x, and hypotenuse 1. By the S Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have r s S cos sin 1 x cos 1 x 2 .

Y

n Y"

Another method: Let u sin 1 x. We need to find cos u in terms of x. To do so, we write cosine in terms of sine. Note S e d

1 x. Now cos u 1 sin2 u is positive because u lies in the interval . that 2 n u n 2 because u sin 2 2 S Substituting u sin 1 x and using the cancellation property sin sin 1 x x gives cos sin 1 x 1 x 2 . r s 36. We want to find sin tan 1 x . Let tan 1 x, so tan x. We sketch a right Y"

triangle with an acute angle , opposite side x, and adjacent side 1. By the S Pythagorean Theorem, the remaining leg is x 2 1. From the figure we have s r x . sin tan 1 x sin S x2 1 r s 37. We want to find tan sin 1 x . Let sin 1 x, so sin x. We sketch a right triangle with an acute angle , opposite side x, and hypotenuse 1. By the S Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have s r x . tan sin 1 x tan S 1 x2 r s 38. We want to find cos tan 1 x . Let tan 1 x, so tan x. We sketch a right

Y n

Y

n Y"

Y"

Y triangle with an acute angle , opposite side x, and adjacent side 1. By the S n Pythagorean Theorem, the remaining leg is x 2 1. From the figure we have

r s 1 1 cos tan x cos S . x2 1 39. Let represent the angle of elevation of the ladder. Let h represent the height, in meters, that the ladder reaches on the

1 0 3 s 1 266 rad s 72 5i . By the Pythagorean Theorem, h 2 1.82 6 2 % building. Then cos 1.8 6 0 3 % cos T T h 36 3.24 32 .76 s 5.7 m.

1 0 8 s 38 9i .

40. Let be the angle of elevation of the sun. Then tan 29 36 0 806 % tan

41. (a) Solving tan h 2 for h, we have h 2 tan . (b) Solving tan h 2 for we have tan 1 h 2 .

42. (a) Solving tan 15 s. Solving for , we have tan 1 15 s .

5 1 5 s 68 2i . (b) Set s 6 m to get tan 15 6 2 . Solving for , we have tan 2

SECTION 6.5 The Law of Sines

493

43. (a) Solving sin h 204 for we have sin 1 h 204 . r s i (b) Set h 1 50 to get sin 1 150 204 s 0 826 rad s 47 3 .

6340 . Solving for , we get 44. (a) Since the radius of the earth is 634 0 k m, we have the relationship cos h 6340 r s 6340 . cos 1 h 6 340

(b) The arc length is s radius included angle 6340 2 1268 0 . r s 6340 (c) s 1268 0 cos 1 h 6 340 r s r s 6340 (d) When h 16 0 we have s 12680 cos 1 160 6340 1268 0 cos 1 6340 6500 s 2819 2 km. r s r s r s 3640 6340 6340 3920 3920 1 (e) When s 392 0 we have 3920 12680 cos 1 h 6 h 6340 % h 6340 cos 12680 % 340 % 12680 cos r s r s 3920 % h 63 4 0 sec 3920 6340 s 315 5 km. h 6340 63 4 0 sec 12680 12680 t u t u 1 1 1 1 45. (a) sin sin s sin 1 0 8102 s 54 1i 7 tan 10i 2 3 1 tan 10i t u t u 1 1 i . For n 3, sin 1 (b) For n 2, sin 1 s 48 3 s 32 2i . For n 4, 5 tan 15i 7 tan 15i t u 1 1 1 s 3 732 s 24 5i . n 0 and n 1 are outside of the domain for 15i , because sin 9 tan 15i tan 15i 1 and s 1 244, neither of which is in the domain of sin 1 .

3 tan 15i 1 46. Let sec 1 x. Then sec x, as shown in the figure. Then cos , so x t u t u Y 1 1 . Thus, sec 1 x cos 1 , x o 1. cos 1 x x n

In particular, sec 1 2 cos 1 21 3.

1 Let csc 1 x. Then csc x, as shown in the figure. Then sin , so x t u t u 1 1 sin 1 . Thus, csc 1 x sin 1 , x o 1. x x

1 Let cot 1 x. Then cot x, as shown in the figure. Then tan , so x t u t u 1 1 tan 1 . Thus, cot 1 x tan 1 , x o 1. x x In particular, cot 1 4 tan 1 41 s 0 245.

a Y

THE LAW OF SINES

sin B sin C sin A

. a b c 2. (a) The Law of Sines can be used to solve triangles in cases ASA or SSA.

1. In triangle ABC with sides a, b, and c the Law of Sines states that

(b) The Law of Sines can give ambiguous solutions in case SSA. 3. / C 180i 98 4i 24 6i 57i . x

z

In particular, csc 1 3 sin 1 31 s 0 340.

6.5

Y

376 sin 57i s 318 75. sin 98 4i

494

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

4. / C 180i 37 5i 28 1i 114 4i . x 5. / C 180i 52i 70i 58i . x

17 sin 114 4i s 25 4. sin 37 5i

26 7 sin 52i s 24 8. sin 58i

6. sin

56 3 sin 67i s 0 646. Then s sin 1 0 646 s 40 3i .

80 2

7. sin C

36 sin 120i s 0 693 % / C s sin 1 0 693 s 44i .

45

8. / C 180i 102i 28i 50i . x

185 sin 50i s 144 9. sin 102i

9. / C 180i 46i 20i 114i . Then a

65 sin 46i 65 sin 20i s 51 and b s 24. i sin 114 sin 114i

10. / B 180i 30i 100i 50i . Then c

2 sin 100i 2 sin 30i s 2 57 and a s 1 31. i sin 50 sin 50i

11. / B 68i , so / A 180i 68i 68i 44i and a 12. sin B

12 sin 44i s 8 99. sin 68i

3 4 sin 80i 6 5 sin 69i s 0 515, so / B sin 1 0 515 s 31i . Then / C 180i 80i 31i 69i and c s 6 2.

6 5 sin 80i

13. / C 180i 50i 68i 62i . Then 230 sin 50i 230 sin 68i a s 200 and b s 242. i sin 62 sin 62i

14. / C 180i 110i 23i 47i . Then 50 sin 23i 50 sin 110i a s 26 7 and b s 64 2. i sin 47 sin 47i %

%

c

c # c

c #

$

$

15. / B 180i 30i 65i 85i . Then 10 sin 65i 10 sin 30i s 5 0 and c s 9. a i sin 85 sin 85i

16. / C 180i 95i 22i 63i . Then 420 sin 63i 420 sin 95i s 1116 9 and c s 999 0. b i sin 22 sin 22i

%

%

c

c

c

c #

#

$

17. / A 180i 51i 29i 100i . Then 44 sin 51i 44 sin 100i s 89 and c s 71. a i sin 29 sin 29i

$

18. / A 180i 100i 10i 70i . Then 115 sin 10i 115 sin 70i s 109 7 and b s 20 3. a i sin 100 sin 100i %

%

c

c c

#

c #

$

$

SECTION 6.5 The Law of Sines

19. Since / A

90i there is only one triangle. sin B

495

15 sin 110i s 0 503 % / B s sin 1 0 503 s 30i . Then 28

28 sin 40i i / i / C s 180i 110i 30i 40i , and so c / i s 19. Thus B s 30 , C s 40 , and c s 19.

sin 110 40 sin 37i 1 / 20. sin C s 0 802 % C1 s sin 0 822 s 53 4i or / C2 s 180i 53 4i s 126 6i .

30 30 sin 89 6i s 49 8.

If / C1 s 53 4i , then / B1 s 180i 37i 53 4i 89 6i and b1 sin 37i 30 sin 16 4i s 14 1.

If / C2 s 126 6i , then / B2 s 180i 37i 126 6i 16 4i and b2 sin 37i Thus, one triangle has / B1 s 89 6i , / C1 s 53 4i , and b1 s 49 8; the other has / B2 s 16 4i , / C2 s 126 6i , and b2 s 14 1.

21. / A 125i is the largest angle, but since side a is not the longest side, there can be no such triangle.

45 sin 38i 22. sin B s 0 660 % / B1 s sin 1 0 660 s 41 3i or / B2 s 180i 41 3i s 138 7i .

42 42 sin 100 7i s 67.

If / B1 s 41 3i , then / A1 s 180i 38i 41 3i 100 7i and a1 sin 38i i 42 sin 3 3 s 3 9.

If / B2 s 138 7i , then / A2 s 180i 38i 138 7i 3 3i and a2 sin 38i i i Thus, one triangle has / A1 s 100 7 , / B1 s 41 3 , and a1 s 67; the other has / A2 s 3 3i , / B2 s 138 7i , and a2 s 3 9.

30 sin 25i 23. sin C s 0 507 % / C1 s sin 1 0 507 s 30 47i or / C2 s 180i 39 47i 149 53i .

25 25 sin 124 53i s 48 73.

If / C1 30 47i , then / A1 s 180i 25i 30 47i 124 53i and a1 sin 25i i 25 sin 5 47 If / C2 149 53i , then / A2 s 180i 25i 149 53i 5 47i and a2 s 5 64. sin 25i Thus, one triangle has / A1 s 125i , / C1 s 30i , and a1 s 49; the other has / A2 s 5i , / C2 s 150i , and a2 s 5 6.

100 sin 30i 24. sin B 23 % / B1 s sin 1 32 s 41 8i or / B2 s 180i 41 8i 138 2i .

75 75 sin 108 2i s 142 5.

If / B1 s 41 8i , then / C1 s 180i 30i 41 8i 108 2i and c1 s sin 30i 75 sin 11 8i If / B2 s 138 2i , then / C2 s 180i 30i 138 2i 11 8i and c2 s s 30 7. sin 30i Thus, one triangle has / B1 s 41 8i , / C1 s 108 2i , and c1 s 142 5; the other has / B2 s 138 2i , / C2 s 11 8i , and c2 s 30 7.

100 sin 50i s 1 532. Since sin n 1 for all , there can be no such angle B, and thus no such triangle.

25. sin B 50 i 80 sin 135 26. sin B s 0 566 % / B1 s sin 1 0 566 s 34 4i or / B2 s 180i 34 4i 145 6i .

100 100 sin 10 6i If / B1 s 34 4i , then / C s 180i 135i 34 4i 10 6i and c s s 25 9. sin 135i If / B2 s 180i 34 4i 145 6i , then / A / B2 135i 145 6i 180i , so there is no such triangle.

Thus, the only possible triangle is / B s 34 4i , / C s 10 6i , and c s 25 9.

26 sin 29i 27. sin A s 0 840 % / A1 s sin 1 0 840 s 57 2i or / A2 s 180i 57 2i 122 8i .

15 15 sin 93 8i s 30 9.

If / A1 s 57 2i , then / B1 180i 29i 57 2i 93 8i and b1 s sin 29i 15 sin 28 1i s 14 6.

If / A2 s 122 8i , then / B2 180 29i 122 8i 28 2i and b2 s sin 29i Thus, one triangle has / A1 s 57 2i , / B1 s 93 8i , and b1 s 30 9; the other has / A2 s 122 8i , / B2 s 28 2i , and b2 s 14 6.

496

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

82 sin 58i s 0 953, so / C1 s sin 1 0 953 s 72 4i or / C2 s 180 72 4i 107 6i .

73 73 sin 49 6i s 65 6.

If / C1 s 72 4i then / A1 s 180i 58i 72 4i 49 6i and a1 s sin 58i 73 sin 14 4i If / C2 s 107 6i , then / A2 s 180i 58i 107 6i 14 4i and a2 s s 21 4. sin 58i i i Thus, one triangle has / A1 s 49 6 , / C1 s 72 4 , and a1 s 65 6; the other has / A2 s 14 4i , / C2 s 107 6i , and a2 s 21 4.

28. sin C

sin B 28 sin 30i sin 30i

% sin B 0 7, so 20 28 20 / B s sin 1 0 7 s 44 427i . Since -BC D is isosceles, / B / B DC s 44 427i . Thus, / BC D 180i 2 / B s 91 146i s 91 1i .

29. (a) From -ABC and the Law of Sines we get

(b) From -ABC we get / BC A 180i / A / B s 180i 30i 44 427i 105 573i . Hence / DC A / BC A / BC D s 105 573i 91 146i 14 4i .

30. By symmetry, / DC B 25i , so / A 180i 25i 50i 105i . Then by the Law of Sines, AD

12 sin 25i s 5 25. sin 105i

31. (a) Let a be the distance from satellite to the tracking station A in kilometers. Then the subtended angle at the satellite is 80 sin 84 2i / C 180i 93i 84 2i 2 8i , and so a s 1629 km. sin 2 8i (b) Let d be the distance above the ground in kilometers. Then d 162 8 8 sin 87i s 1627 km.

32. (a) Let x be the distance from the plane to point A. Then x 8

sin 48i x sin 48i

%

i i i AB sin 180 32 48 sin 100i

sin 48i s 6 03 km. sin 100i

(b) Let h be the height of the plane. Then sin 32i

h " h 6 0 3 sin 32 i s 3 20 km. x

33. / C 180i 82i 52i 46i , so by the Law of Sines, have AC

200 sin 52i s 219 m. sin 46i

AC AB AB sin 52i

% AC , so substituting we i i sin 52 sin 46 sin 46i

312 sin 48 6i s 0 444 % / ABC s sin 1 0 444 s 26 4i , and so / BC A s 180i 48 6i 26 4i 105i .

527 527 sin 105i s 678 5 m.

Then the distance between A and B is AB sin 48 6i

34. sin / ABC

35.

We draw a diagram. A is the position of the tourist and C is the top of the tower.

%

/ B 90i 5 6i 84 4i and so / C 180i 29 2i 84 4i 66 4i . Thus, by 105 sin 29 2i

c

the Law of Sines, the length of the tower is BC

sin 66 4i

s 55 9 m.

c $

#

36.

The situation is illustrated in the diagram.

%

AC 49. 5 sin 67i s 4 5 57 m , so using the Pythagorean Theorem S we can calculate B A 49.52 45 572 s 19 3m and S AD 54 2 45 57 2 s 29 m. Thus the anchor points are

c $

#

&

B A AD s 19 3 29 48.3 m apart.

SECTION 6.5 The Law of Sines

497

37. The angle subtended by the top of the tree and the sun’s rays is / A 180i 90i 52i 38i . Thus the height of the tree 65 sin 30 i s 53 m. is h sin 38i 38.

Let x be the length of the wire, as shown in the figure. Since 12i , other angles

%

in ABC are 90i 58i 148i , and 180i 12i 148i 20i . a

Thus,

100 sin 148i x

% x

100

s 155 m. sin 148i sin 20i sin 20i

Y z $

`

# c

39. Call the balloon’s position R. Then in -P Q R, we see that / P 62i 32i 30i , and / Q 180i 71i 32i 141i .

QR PQ sin 30i

% Q R 60 s 192 m.

Therefore, / R 180i 30i 141i 9i . So by the Law of Sines, i i sin 30 sin 9 sin 9i 40.

$ z

%

c

Label the diagram as shown, and let the hill’s angle of elevation be . Then applying the Law of Sines to ABC,

sin sin 8i

% 120 30

sin 4 sin 8i s 0 55669 " s sin 1 0 55669 s 33 8i . But from -AB D, / B AD / B 8i 90i , so s 90i 8i 33 8i 48 2i .

` #

&

41. Let d be the distance from the earth to Venus, and let be the angle formed by sun, Venus, and earth. By the Law sin 39 4i sin

s 0 878, so either s sin 1 0 878 s 61 4i or s 180i sin 1 0 878 s 118 6i . of Sines, 1 0 723 0 723 d % d s 1 119 AU; in the second case,

In the first case, sin 180i 39 4i 61 4i sin 39 4i d 0 723 % d s 0 427 AU.

sin 180i 39 4i 118 6i sin 39 4i sin 60i b sin 60i sin B

or sin B . Similarly, applying the Law of b c c i i sin B sin 120 r sin 120 r b Sines to -BC D gives

or sin B . Since sin 120i sin 60i , we have % r c d c d c c d c sin D sin 60i b sin 60i b

(`). Similarly, from -ADC and the Law of Sines we have

or sin D , and r c d b d d a sin 120i b sin 60i a sin 120i b d from -B DC we have sin D . Thus,

%

.

Combining this with c d d c d a c d b c d c d b b a b r a b

1. Solving for r, we find 1 % % (`), we get r a c d c d c d r a a b a b ab . r a b

42. (a) Applying the Law of Sines to -ABC, we get

498

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

4 3 12 cm.

4 3 (c) If a b, then r is infinite, and so the face is a flat disk.

(b) r

43. By the area formula from Section 6.3, the area of -ABC is A 12 ab sin C. Because we are given a and the three sin A a sin B sin B

%b . Thus, angles, we need to find b in terms of these quantities. By the Law of Sines, b a sin A u t a 2 sin B sin C a sin B A 12 ab sin C 12 a sin C . sin A 2 sin A 44. By the area formula from Section 6.3,

1 ab sin C sin C Area of - ABC 12

, because a and b are the same for both ) ) ) Area of - A B C sin C ) ab sin C ) 2

triangles.

45.

% C

B

% $

C

$

a o b: One solution

b

B

% C

B{

#

$

a

b sin A: Two solutions

$

#

% C

B $

#

a b sin A: One solution

a b sin A: No solution

/ A 30i , b 100, sin A 1 . If a o b 100 then there is one triangle. If 100 2

a 100 sin 30i 50, then there are two possible triangles. If a 50, then there is one (right) triangle. And if a 50, then no triangle is possible.

6.6

THE LAW OF COSINES

1. For triangle ABC with sides a, b, and c the Law of Cosines states c2 a 2 b2 2ab cos C. 2. The Law of Cosines is required to solve triangles in cases SSS and SAS.

T 834 115 s 28 9.

T 2 2 2 i i 4. x 15 18 2 15 18 cos 108 225 324 540 cos 108 s 715 869 and so x s 715 869 s 26 8.

T 5. x 2 252 252 2 25 25 cos 140i 625 625 1250 cos 140i s 2207 556 and so x s 2207 556 s 47.

T 6. x 2 22 82 2 2 8 cos 88i 4 64 32 cos 88i s 66 883 and so x s 66 883 s 8 2.

3. x 2 212 422 2 21 42 cos 39i 441 1764 1764 cos 39i s 834 115 and so x s

7. 37 832 68 012 42 152 2 68 01 42 15 cos . Then cos

37 832 68 012 42 152 s 0 867 % 2 68 01 42 15

s cos 1 0 867 s 29 89i .

8. 154 62 60 12 122 52 2 60 1 122 5 cos . Then cos

154 62 60 12 122 52 s 0 359 % 2 60 1 122 5

s cos 1 0 359 s 111i .

9. x 2 242 302 2 24 30 cos 30i 576 900 1440 cos 30i s 228 923 and so x s

T 228 923 s 15.

156 202 102 122

0 65 % s cos 1 0 65 s 130 54i .

2 10 12 240 T 11. c2 102 182 2 10 18 cos 120i 100 324 360 cos 120i 604 and so c s 604 s 24 576. Then 18 sin 120i sin A s s 0 634295 % / A s sin 1 0 634295 s 39 4i , and / B s 180i 120i 39 4i 20 6i .

24 576

10. 202 102 122 2 10 12 cos . Then cos

12. 122 402 442 2 40 44 cos B % cos B sin A

122 402 442 s 0 964 % / B s cos 1 0 964 s 15i . Then 2 40 44

40 sin 15 5i s 0 891 % / A s sin 1 0 891 s 63i , and so / C s 180i 15i 63i 102i .

12

SECTION 6.6 The Law of Cosines

13. c2 32 42 2 3 4 cos 53i 9 16 24 cos 53i s 10 556 % c s

499

T 4 sin 53i s 0 983 10 556 s 3 2. Then sin B 3 25

% / B s sin 1 0 983 s 79i and / A s 180i 53i 79i 48i .

14. a 2 602 302 2 60 30 cos 70i 3600 900 3600 cos 70i s 3268 73 T 30 sin 70i s 0 493 % % a 3268 73 s 57 2. Then sin C s 57 2 / C s sin 1 0 493 s 29 5i , and / B s 180i 70i 29 5i 80 5i .

2

2

2

25 22 s 0 644 % / A s cos 1 0 644 s 50i . Then 15. 202 252 222 2 25 22 cos A % cos A 20 2 25 22 i

sin B s 25 sin2049 9 s 0 956 % / B s sin 1 0 956 s 73i , and so / C s 180i 50i 73i 57i .

2

2

2

12 16 0 78125 % 16. 102 122 162 2 12 16 cos A % cos A 10 2 12 16 i / A s cos 1 0 78125 s 38 6i . Then sin B s 12 sin 38 6 s 0 749 % 10

/ B s sin 1 0 749 s 48 5i , and so / C s 180i 38 6i 48 5i 92 9i . i

sin 40 s 0 833 % / C s sin 1 0 833 s 56 4i or / C s 180i 56 4i s 123 6i .

17. sin C 162125 1 2 i

sin 83 6 s 193.

If / C1 s 56 4i , then / A1 s 180i 40i 56 4i 83 6i and a1 125sin 40i i

sin 16 4 s 54 9.

If / C2 s 123 6i , then / A2 s 180i 40i 123 6i 16 4i and a2 125sin 40i

Thus, one triangle has / A s 83 6i , / C s 56 4i , and a s 193; the other has / A s 16 4i , / C s 123 6i , and a s 54 9. i

52 s 1 024. Since sin n 1 for all , there is no such / A, and hence there is no such triangle.

18. sin A 65 sin 50 i

55 s 1 065. Since sin n 1 for all , there is no such / B, and hence there is no such triangle.

19. sin B 65 sin 50 i

i

sin 61 s 109 4 and c 73 5 sin 83 s 124 1.

20. / A 180i 61i 83i 36i . Then b 73 5 sin 36i sin 36i i

sin 35 s 2.

21. / B 180i 35i 85i 60i . Then x 3sin 60i

22. x 2 10i 18i 2 10 18 cos 40i 100 324 360 cos 40i s 148 224 and so x s

T 148 224 s 12 2.

i

sin 30 23. x 50 sin 100i s 25 4 2

2

2

10 11 205 s 0 932 % s cos 1 0 932 s 21 3i .

24. 42 102 112 2 10 11 cos . Then cos 4 2 10 11 220

25. b2 1102 1382 2 110 138 cos 38i 12,100 19,044 30,360 cos 38i s 7220 0 and so b s 85 0. Therefore, 2

2

85 128 using the Law of Cosines again, we have cos 1102 110 138

2

% s 89 15i .

i

40 s 0 803 % s sin 1 0 803 s 53 5i or s 180i 53 5i s 126 5i , but 53 5i doesn’t fit the 26. sin 10 sin 8

picture, so s 126 5i .

27. x 2 382 482 2 38 48 cos 30i 1444 2304 3648 cos 30i s 588 739 and so x s 24 3. i

sin 98 s 1180 8.

28. / A 180i 98i 25i 57i . Then x 1000 sin 57i

18, so by Heron’s Formula the area is 29. The semiperimeter is s 9 12 15 2 T T A 18 18 9 18 12 18 15 2916 54.

52 , so by Heron’s Formula the area is 30. The semiperimeter is s 1 2 2 2 U r T T sr sr s 15 A 52 52 1 52 2 52 2 15 16 4 s 0 968.

12, so by Heron’s Formula the area is 31. The semiperimeter is s 7 8 9 2 T T T A 12 12 7 12 8 12 9 720 12 5 s 26 8.

32. The semiperimeter is s 11 100 101 106, so by Heron’s Formula the area is 2 T T T A 106 106 11 106 100 106 101 302,100 10 3021 s 550.

500

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

33. The semiperimeter is s 3 4 6 13 2 2 , so by Heron’s Formula the area is U r T sr sr s T 13 3 13 4 13 6 455 455 s 5 33.

A 13 2 2 2 2 16 4 6, so by Heron’s Formula the 34. Both of the smaller triangles have the same area. The semiperimeter of each is s 2 5 5 2 T T T T area of each is A 6 6 2 6 5 6 5 24 2 6, so the shaded area is 4 6 s 9 80.

35. We draw a diagonal connecting the vertices adjacent to the 100i angle. This forms two triangles. Consider the triangle with sides of length 5 and 6 containing the 100i angle. The area of this triangle is A1 12 5 6 sin 100i s 14 77. To use Heron’s Formula to find the area of the second triangle, we need to find the length of the diagonal using the Law of Cosines: c2 a 2 b2 2ab cos C 52 62 2 5 6 cos 100i s 71 419 " c s 8 45. Thus the second triangle has semiperimeter T 8 7 8 45 s 11 7255 and area A2 11 7255 11 7255 8 11 7255 7 11 7255 8 45 s 26 00. The area s 2 of the quadrilateral is the sum of the areas of the two triangles: A A1 A2 s 14 77 26 00 40 77.

36. We draw a line segment with length x bisecting the 60i angle to create two triangles. By the Law of Cosines, T T T 32 42 x 2 2 4x cos 30i % x 2 4 3x 7 0. Using the Quadratic Formula, we find x 2 3 5. The minus sign provides the correct length of about 1 23 (the other solution is about 5 7, which corresponds to a convex quadrilateral T T 3 4 2 3 5 with the same side lengths), so the semiperimeter of each triangle is s and the total area of the figure 2 U r T T s is A 2 s s 3 s 4 s 2 3 5 s 2 46.

Label the centers of the circles A, B, and C, as in the figure. By the Law of

37. %

$

#

Cosines, cos A

92 102 112 AB 2 AC 2 BC 2

13 " / A s 70 53i .

2 AB AC 2 9 10

Now, by the Law of Sines,

sin 70 53i sin B sin C

. So 11 AC AB

i 1 0 85710 s 58 99i and sin B 10 11 sin 70 53 s 0 85710 " B s sin 9 sin 70 53i s 0 77139 " C s sin 1 0 77139 s 50 48i . The area of sin C 11

ABC is 12 AB AC sin A 12 9 10 sin 70 53i s 42 426. i

2 70 53 s 9 848. Similarly, the areas of sectors B and C

4 360i 360i are S B s 12 870 and SC s 15 859. Thus, the area enclosed between the circles is A ABC S A S B SC "

The area of sector A is given by S A R 2

A s 42 426 9 848 12 870 15 859 s 3 85 cm2 .

T 38. By the Law of Cosines we have a 2 62 42 2 6 4 cos 45i 52 24 2, rT T T s 2 6 x 16. b2 62 x 2 2 6 x cos 30i x 2 6 3x 36, and c2 x 2 42 2 x 4 cos 30i 45i x 2 2 r s rT T T T s By the Pythagorean Theorem, a 2 b2 c2 , so we have 52 24 2 x 2 6 3x 36 x 2 2 2 6 x 16 r T s rT r T 12 3 2 T T T T T s T s % 26 12 2 3 3x 18 2 6 x 8 % 36 12 2 3 3 2 6 x, so x T T T .

3 3 2 6 39. Let c be the distance across the lake, in kilometers. Then c2 2 822 3 562 2 2 82 3 56 cos 40 3i s 5 313 % c s 2 30 km.

SECTION 6.6 The Law of Cosines &

40. Suppose ABC D is a parallelogram with AB DC 5, AD BC 3,

%

and / A 50i (see the figure). Since opposite angles are equal in a

parallelogram, it follows that / C 50i , and 260i / B / D 360i 100i 260i . Thus, / B / D 130i .

501

cM #

$

2 T By the Law of Cosines, AC 2 32 52 2 3 5 cos 130i " AC 9 25 30 cos 130i s 7 3. Similarly, S B D 32 52 2 3 5 cos 50i s 3 8.

41. In half an hour, the faster car travels 25 kilometers while the slower car travels 15 kilometers. The distance between them is given by the Law of Cosines: d 2 252 152 2 25 15 cos 65i " S T d 252 152 2 25 15 cos 65i 5 25 9 30 cos 65i s 23 1 km.

42. Let x be the car’s distance from its original position. Since the car travels Y

at a constant speed of 40 kilometers per hour, it must have traveled 40 kilometers

east, and then 20 kilometers northeast (which is 45i east of “due north”). From

z c

the diagram, we see that / 135i , so S T x 202 402 2 20 40 cos 135i 10 4 16 16 cos 135i s 56 0 k m.

43. The pilot travels a distance of 1000 1 5 1500 kilometers in her original direction and 1000 2 200 0 kilometer s in the new direction. Since she makes a course correction of 10 i to the right, the included angle is

c

E

180i 10i 170i . From the figure, we use the Law of Cosines to get the expression d 2 1500 2 2000 2 2 1500 2000 cos 170i s 12,158,846 52, so d s 3487 kilometers. Thus, the pilot’s distance from her original position is approximately 348 7 k m.

44. Let d be the distance between the two boats in kilometers. After one hour, the boats have traveled distances of 30 km and 26 km. Also, the angle subtended by their directions is 180i 50i 70i 60i . Then T d 2 302 262 2 30 26 cos 60i 796 % d s 796 s 28 2. Thus the distance between the two boats is about 28 kilometers.

45. (a) The angle subtended at Egg Island is 100i . Thus using the Law of

Forrest Island

Cosines, the distance from Forrest Island to the fisherman’s home port is x 2 302 502 2 30 50 cos 100i 900 2500 3000 cos 100i s 3920 945

c

T and so x s 3920 945 s 62 62 kilometers.

Y

(b) Let be the angle shown in the figure. Using the Law of Sines, 50 sin 100i s 0 7863 % s sin 1 0 7863 s 51 8i . Then sin 62 62 90i 20i 51 8i 18 2i . Thus the bearing to his home port is S 18 2i E.

c Egg Island

c

a n c Home Port

502

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

46. (a) In 30 minutes the pilot flies 160 kilometers due east, so using the Law of

$

Cosines we have x 2 1602 4802 2 160 480 cos 40 i

1602 1 9 6 cos 40i s 1602 5 404 .

S Thus, x s 1602 5 404 s 371 9, and so the pilot is 371 9 kilometers from his destination. (b) Using the Law of Sines, sin

#

c c

Y

n

480 sin 40 i s 0 829 % 371 9

s sin 1 0 829 s 56i . However, since

90i , the angle we seek is

180i 56i 124i . Hence the bearing is 124i 90i 34i , that is, N 34i E. 47. The largest angle is the one opposite the longest side; call this angle . Then by the Law of Cosines, 442 362 222 2 36 22 cos % cos

362 222 442 0 09848 " s cos 1 0 09848 s 96i .

2 36 22

48. Let be the angle formed by the cables. The two tugboats and the barge form a triangle: the side opposite has a length of

36 m and the other two sides have lengths of 64 and 69 m. Therefore, 36 2 64 2 69 2 2 64 69 cos % cos

64 2 69 2 36 2 " cos 0 856 1 " s cos 1 0 8561s 31 i.

2 64 69

49. Let d be the distance between the kites. Then d 2 s 1142 1262 2 114 126 cos 30i " S d s 1142 1262 2 114 126 cos 30i s 63 m.

50. Let x be the length of the wire and the angle opposite x, as shown in the figure.

Since the mountain is inclined 32i , we must have 180i 90i 32i 122i . Thus, S x 172 17 2 2 17 38 cos 122i s 49 m.

Y n

c

51. Solution 1: From the figure, we see that 106i and sin 74i

1020 b

%

1020 s 1061. Thus, x2 2402 10612 2 240 1061 cos 106i sin 74i S x 2402 10612 2 240 1061 cos 106i " x s 1151 m .

b

Solution 2: Notice that tan 74i

1020 a

%

a

" Y C

1020 s 292 5. By the tan 74i

Pythagorean Theorem, x 2 a 2 4 0 2 10202 . So T x 292 5 240 2 10202 s 1151 m.

a

c

52. Let the woman be at point A, the first landmark (at 62i ) be at point B, and the other landmark be at point C. We 345 345 345 % AB % s 735. Similarly, cos 54 i want to find the length BC. Now, cos 62i AB cos 62i AC 345 s 587. Therefore, by the Law of Cosines, BC 2 AB 2 AC 2 2 AB AC cos 43i " AC cos 54i S BC 735 2 587 2 2 735 587 cos 43i " BC s 504. Thus, the two landmarks are roughly 504 meters apart.

CHAPTER 6

Review

503

T a b c 34 44 57

67.5. Thus, 53. By Heron’s formula, A s s a s b s c , where s 2 2 T A 67.5 67.5 34 6 7 . 5 44 67.5 57 s 747 m 2 . Since the land value is $20 0 per square meter , the value of the lot is approximately 74 7 20 0 $149,400.

54. Having found a s 13 2 using the Law of Cosines, we use the Law of Sines to find / B:

sin 46 5i sin B

% 10 5 13 2

10 5 sin 46 5i s 0 577. Now there are two angles / B between 0i and 180i which have sin B 0 577, namely 13 2 / B s 35 2i and / B s 144 8i . But we must choose / B , since otherwise / A / B 180i .

1 2 1 i i sin 46 5 18 0 sin 46 5 sin C

% sin C s 0 989, so either / C s 81 5i Using the Law of Sines again, 18 0 13 2 13 2 or / C s 98 5i . In this case we must choose / C s 98 5i so that the sum of the angles in the triangle is / A / B / C s 46 5i 35 2i 98 5i s 180i . (The fact that the angles do not sum to exactly 180i , and the discrepancies between these results and those of Example 3, are due to roundoff error.)

The method in this exercise is slightly easier computationally, but the method in Example 3 is more reliable. sin B

55. In any ABC, the Law of Cosines gives a 2 b2 c2 2bc cos A, b2 a 2 c2 2ac cos B, and c2 a 2 b2 2ab cos C. Adding the second and third equations gives b2 a 2 c2 2ac cos B c2 a 2 b2 2ab cos C b2 c2 2a 2 b2 c2 2a c cos B b cos C Thus 2a 2 2a c cos B b cos C 0, and so 2a a c cos B b cos C 0.

Since a / 0 we must have a c cos B b cos C 0 % a b cos C c cos B . The other laws follow from the symmetry of a, b, and c.

CHAPTER 6 REVIEW s 0 52 rad 1. (a) 30i 30 180 6

7 s 1 83 rad 2. (a) 105i 105 180 12

5 s 2 62 rad (b) 150i 150 180 6

2 s 1 26 rad (b) 72i 72 180 5

s 0 35 rad (c) 20i 20 180 9

9 s 7 07 rad (c) 405i 405 180 4

5 s 3 93 rad (d) 225i 225 180 4

7 s 5 50 rad (d) 315i 315 180 4

i 3. (a) 56 rad 56 180 150

i 4. (a) 53 rad 53 180 300

180 i (b) 9 rad 9 20

i (b) 109 rad 109 180 200

i (c) 43 rad 43 180 240

900 i (c) 5 rad 5 180 s 286 5

720 i (d) 4 rad 4 180 s 229 2

i (d) 113 rad 113 180 660

5. r 10 m, 25 rad. Then s r 10 25 4 s 12 6 m.

6. s 7 cm, r 2 5 cm. Then

s 7 2 8 rad s 160 4i 2 5 r

5 rad. Then r s 25 18 90 s 28 6 m.

7. s 25 m , 50 i 50 180

18 5

13 rad. Then r s 13 18 18 m.

8. s 13 m, 130i 130 180 18 13

504

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

9. Since the diameter is 70 cm , r 35cm . In one revolution, the arc length (distance traveled) is s r 2 35 70 cm.

The total distance traveled is 9 6 k m/h 0 5 h 48 k m 48 k m 10 0 0 m /k m 100 cm /m 4 ,8 00,0 00. The number of 1 rev revolution is 4 ,8 00,0 00 in s 21,827 rev. Therefore the car wheel will make approximately 21,827 revolutions.

70 cm 10. r 6340 kilometers, s 3920 kilometers. Then is approximately 35i 4 .

3920 s s 35 448i and so the angle

s 0 619 rad 0 619 180 6340 r

11. r 5 m, 2 rad. Then A 12 r 2 12 52 2 25 m2 . b c s 18,151 m2 12. A 12 r 2 12 200 2 52i 180 i 2A 2 125 i 13. A 125 m 2 , r 25 m . Then 2 250 625 0 4 rad s 22 9 r 252 V U 2A 100 11 s 4 2 m.

14. A 50 m2 and 116 rad. Thus, r

6

150 2 rad 300 rad/min s 942 5 rad/min. The linear speed is 1 min

15. The angular speed is $ )

150 2 2 0 6000 cm /min s 18,849 6 cm /min s 188 5 m /min.

1

rad 7000 rad/min s 21,991 1 rad/min.

16. (a) The angular speed of the engine is $e 3500 2 1 min $e 7000 $rad/min 0 9 % (b) To find the angular speed $* of the wheels, we calculate g $ * *

$* s 7777 8 rad/min s 24,434 6 rad/min. (c) The speed of the car is the angular speed of the wheels times their radius: 7777 8 rad 28 cm 60 min 1 k m min 1 hr 100,000 cm s 410 5 k m/h.

17. r

T T S T 52 72 74. Then sin T5 , cos T7 , tan 57 , csc 574 , sec 774 , and cot 75 .

74

74

T T S T 3 , cos 91 , tan T3 , csc 10 , sec T10 , and cot 91 .

18. x 102 32 91. Then sin 10 10 3 3 91

x 19. cos 40i 5 20. cos 35i

%

x 5 cos 40i s 3 83, and 5y sin 40i

%

91

y 5 sin 40i s 3 21.

2 % x cos235i s 2 44, and tan 35i 2y % y 2 tan 35i s 1 40. x

1 2 924 s 3 11. sin 20i % x sin120i s 2 92, and xy cos 20i % y cosx20i s 0 9397 x x 22. cos 30i % x 4 cos 30i s 3 46, and sin 30i xy % y x sin 30i 3 46 0 5 s 1 73.

4

21.

23. A 90i 20i 70i , a 3 cos 20i s 2 819, and b 3 sin 20i s 1 026.

a 20 cos 60i 40, and tan 60i b 20 % b 20 tan 60i s 34 64.

%

C #

24. C 90i 60i 30i , cos 60i 20 a %

%

c

B

c

C $

c

B c

#

$

CHAPTER 6

25. c

S 7 s 0 2960 s 17 0i , 252 72 24, A sin 1 24

26. b

i and C sin 1 24 25 s 1 2870 s 73 7 .

#

D

505

S 5 s 0 3948 s 22 6i , 122 52 13, A sin 1 13

i and C sin 1 12 13 s 1 1760 s 67 4 . %

Review

% C

$

#

$

1 1 1 1 %a cot , sin % b csc a tan b sin h 28. Let h be the height of the tower in meters. Then tan 28 81i % h 1000 tan 28 81i s 550 m.

1000

27. tan

One side of the hexagon together with radial line segments through its endpoints

29.

Y

forms a triangle with two sides of length 8 m and subtended angle 60i . Let x be the

c

length of one such side (in meters). By the Law of Cosines,

x 2 82 82 2 8 8 cos 60i 64

%

x 8. Thus the perimeter of the

hexagon is 6x 6 8 48 m.

30. As the crankshaft moves in its circular pattern, point Q is

Z

Z

determined by the angle , namely it has coordinates Q 5 cos , 5 sin . We split the triangle into two right triangles -O Q R and -P Q R, as shown in the figure. Let h be the height of the piston. We consider two cases, 0i n 180i and

I

I

180i n 360i .

If 0i n 180i , then h is the sum of O R and R P. Using the T Pythagorean Theorem, we find R P 20 2 5 cos 2 , while

4 n 1

3 cos n sin n n 1

Y 3 cos n s in n

Y

4

O R is the y-coordinate of the point Q, 2 sin . Thus S h 400 25 cos 2 5 sin .

S If 180i n 360i , then h is the difference between R P and R O. Again, R P 400 25 cos 2 and O R is the S y-coordinate of the point Q, 2 sin . Thus h 400 25 cos 2 5 sin . Since sin 0 for 180i 360i , this also S reduces to h 400 25 cos 2 5 sin .

S Since we get the same result in both cases, the height of the piston in terms of is h 400 25 cos 2 5 sin .

r

, 0 518i % r r 379,040 tan 0 259i 2 r AB 379,040 tan 0 259i % r 1 tan 0 259i 379,040 tan 0 259i % r s 1722 and so the radius of the moon is 1 tan 0 259i roughly 1722 kilometers.

31. Let r represent the radius, in kilo meters , of the moon. Then tan

32. Let d1 represent the horizontal distance from a point directly below the plane to the closer ship in meters , and d2 represent the 10,500 10,500 10,500 % d1 , and similarly tan 40i % horizontal distance to the other ship in meters . Then tan 52i d1 tan 52i d2 10,500 10,500 10,500 d2 . So the distance between the two ships is d2 d1

s 4320 m . tan 40i tan 40i tan 52i T T 34. csc 94 csc 33. sin 315i sin 45i T1 22 4 2 2

35. tan 135i tan 45i 1

T

3 36. cos 56 cos 6 2

506

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

T s r 3 T1 37. cot 223 cot 23 cot 3 3 3

38. sin 405i sin 45i T1 22 2

39. cos 585i cos 225i cos 45i T1 22

40. sec 223 sec 43 sec 3 2

T

T

T

2

T

2 3 T2 41. csc 83 csc 23 csc 3 3 3

2 3 42. sec 136 sec 6 3

T 43. cot 390i cot 30i cot 30i 3 44. tan 234 tan 34 tan 4 1 T T 5 12 13 13 45. r 5 2 122 169 13. Then sin 12 13 , cos 13 , tan 5 , csc 12 , sec 5 , and 5 . cot 12

46. If is in standard position, then the terminal point of on the unit circle is simply cos sin . Since the terminal point is s r T given as 23 12 , sin 12 .

T T T T 47. y 3x 1 0 % y 3x 1, so the slope of the line is m 3. Then tan m 3 % 60i .

S T 48. 4y 2x 1 0 % y 12 x 14 . The slope of the line is m 12 . Then tan m 12 and r 12 22 5. So T T sin T1 , cos T2 , tan 12 , csc 5, sec 25 , and cot 2.

5 5 S S sin 1 cos2 49. Since sin is positive in quadrant II, sin 1 cos2 and we have tan

. cos cos 1 1 50. sec S (because cos 0 in quadrant III). cos 1 sin2 51. tan2

sin2 sin2

cos2 1 sin2

52. csc2 cos2

1 sin2

cos2

1 sin2 sin2

1

1

sin2 T T T T 53. tan 37 , sec 43 . Then cos 34 and sin tan cos 47 , csc T4 4 7 7 , and cot T3 3 7 7 .

7 7 9 sin 41 9 , and cot 40 .

40 40 9 cos 41 S T 55. sin 35 . Since cos 0, is in quadrant II. Thus, x 52 32 16 4 and so cos 45 , tan 34 , 41 9 40 54. sec 41 40 , csc 9 . Then sin 41 , cos 41 , tan

csc 53 , sec 54 , cot 43 .

5 , and must be in quadrant III " sin 0. Therefore, 56. sec 13 0. Then cos 13 5 and tan T S 25 12 , csc 13 , tan sin 12 , and cot 5 . sin 1 cos2 1 169 13 12 5 12 cos T 4 " cos 45 T2 since 57. tan 12 . sec2 1 tan2 1 14 54 % cos2 5 5 r s sin 1 1 1 2 1 2 % sin 2 cos 2 T T . Therefore, cos 0 in quadrant II. But tan 5 5 cos T s r 5 1 2 1 sin cos T T T 5 .

5 5 5 T 1 1 1 sin 1 sin 1 sin 2 3 T 58. sin 12 for in quadrant I. Then tan sec

S 3.

T2 cos cos cos 2 1 sin2 1 1 2

59. By the Pythagorean Theorem, sin2 cos2 1 for any angle .

T

T

3 5 10 5 5

60. cos 23 and 2 . Then 6 " 2 6 3 . So sin 2 sin 3 sin 3 2 .

T

61. sin 1 23 3

T

62. tan 1 33 6

CHAPTER 6

Review

507

63. Let u sin 1 52 and so sin u 25 . Then from the triangle, 64. Let u cos 1 83 then cos u 38 . From the triangle, we s r T r s tan sin 1 52 tan u T2 . have sin cos 1 83 sin u 855 .

21

V

V

65. Let tan 1 x

% tan x. Then from the r s x triangle, we have sin tan 1 x sin S .

1 x2

66. Let sin 1 x. Then sin x. From the triangle, we r s 1 have sec sin 1 x sec S .

1 x2

Y"

Y

n

n

Y"

rx s x " tan 1 2 2 i 10 sin 30 i i i i s 5 32.

69. / B 180 30 80 70 , and so by the Law of Sines, x sin 70i 2 sin 45i s 1 46 70. x sin 105i 67. cos

rx s x " cos 1 3 3

Y

68. tan

71. x 2 1002 2102 2 100 210 cos 40i s 21,926 133 % x s 148 07 20 sin 60i 72. sin B s 0 247 % / B s sin 1 0 247 s 14 33i . Then / C s 180i 60i 14 33i 105 67i , and so 70 70 sin 105 67i xs s 77 82. sin 60i T 73. x 2 22 82 2 2 8 cos 120i 84 % x s 84 s 9 17 6 sin 31 21i 4 sin 110i s 0 626 % / B s 38 79i . Then / C s 180i 110i 38 79i 31 21i , and so x s s 3 3.

74. sin B 6 sin 110i t u sin 25i 23 sin 25i sin 23 sin 25i

" sin " sin 1 s 54 1i or 75. By the Law of Sines, 23 12 12 12 s 180i 54 1i 125 9i . t u sin sin 80i 4 sin 80i 4 sin 80i 76. By the Law of Sines,

" sin " sin 1 s 52 0i .

4 5 5 5 77. By the Law of Cosines, 1202 1002 852 2 100 85 cos , so cos

1202 1002 852 s 0 16618. Thus, 2 100 85

s cos 1 0 16618 s 80 4i . sin 10i 5 sin 10i sin A

" sin A s 0 2894, so A s sin 1 0 2894 s 16 8i 5 3 3 or A s 180i 16 8i 163 2i . Therefore, s 180i 10i 16 8i 153 2i or 180i 10i 163 2i 6 8i .

78. We first use the Law of Sines to find / A:

79. After 2 hours the ships have traveled distances d1 40 mi and d2 56 km. The subtended angle is 180i 32i 42i 106i . Let d be the distance between the two ships in kilo meter s. Then by the Law of Cosines, d 2 402 562 2 40 56 cos 106i s 5970 855 % d s 77 3 kilo meter s.

508

CHAPTER 6 Trigonometric Functions: Right Triangle Approach

80. Let h represent the height of the building in meters , and x the horizontal distance from the building to point B. Then h h h tan 24 1i and tan 30 2i % x h cot 30 2i . Substituting for x gives tan 24 1i % x 18 0 x h cot 30 2i 18 0 i 18 0 tan 24 1 h tan 24 1i h cot 30 2i 18 0 % h s 348 m .

1 tan 24 1i cot 30 2i 81. Let d be the distance, in kilometers, between the points A and B . Then by the Law of Cosines, d 2 3 22 5 62 2 3 2 5 6 cos 42i s 14 966 % d s 3 9 k m.

120 sin 68 9i s 120 08 kilo meters . Let d be the shortest distance, sin 68 8i in kilometers, to the shore. Then d b sin A s 120 08 sin 42 3i s 80 8 kilometers .

82. / C 180i 42 3i 68 9i 68 8i . Then b 83. A 12 ab sin 12 8 14 sin 35i s 32 12

T 5 6 8 a b c

9 5. Thus, 84. By Heron’s Formula, A s s a s b s c , where s 2 2 T A 9 5 9 5 5 9 5 6 9 5 8 s 14 98.

CHAPTER 6 TEST 11 rad. 135i 135 3 rad.

1. 330i 330 180 6 180 4 180 234 i i 2. 43 rad 43 180 240 . 1 3 rad 1 3 s 74 5

3. (a) The angular speed is $

120 2 rad 240 rad/min s 753 98 rad/min.

1 min

(b) The linear speed is 120 2 5 1200 m /min 1 s 3769 9 m /min

)

T

4. (a) sin 405i sin 45i T1 22 2

T

(b) tan 150i tan 30i T1 33 3 (c) sec 53 sec 3 2 (d) csc 52 csc 2 1

s 226 k m/h rT s T 2 13 3 S T 2 2 26 6 13 .

5. r 32 22 13. Then tan sin T

T 3 39 13 3 13 b a % a 24 sin . Also, cos % b 24 cos .

6. sin 24 24 S T 7. cos 13 and is in quadrant III, so r 3, x 1, and y 32 12 2 2. Then

T T 1 2 3 4 3 2.

3 2 T csc 1 T 4 2 2 2 2 tan 1 sin

1 5 13 13 1 5 , tan 5 . Then sec 8. sin 13

tan

.

12 cos cos sin sin 12 5 12 S S 9. sec2 1 tan2 % tan sec2 1. Thus, tan sec2 1 since tan 0 in quadrant II.

tan cot csc tan

h " h 1.8 tan 73i s 5 9 m .

1.8 rx s x 11. (a) tan " tan 1 4 4 t u 3 3 (b) cos " cos 1 x x 10. tan 73i

Surveying 9 so tan u 9 . From the triangle, r 12. Let u tan 1 40 40 s r 40 9 1 cos tan 40 cos u 41 .

S 92 402 41. So

509

V

13. By the Law of Cosines, x 2 102 122 2 10 12 cos 48i s 8 409 % x s 9 1.

14. / C 180i 52i 69i 59i . Then by the Law of Sines, x

230 sin 69i s 250 5. sin 59i

h x h % h 50 tan 20i and tan 28i % x h 50 tan 28i 50 50 % x 50 tan 28i h 50 tan 28i 50 tan 20i s 8 4.

15 sin 108i s 0 509 % / A s 30 63i . Then 16. Let / A and / X be the other angles in the triangle. Then sin A 28 28 sin 41 37i / X s 180i 108i 30 63i s 41 37i , and so x s s 19 5. sin 108i 15. Let h be the height of the shorter altitude. Then tan 20i

17. By the Law of Cosines, 92 82 62 2 8 6 cos " cos

82 62 92 s 0 1979, so s cos 1 0 1979 s 78 6i .

2 8 6

18. We find the length of the third side x using the Law of Cosines: x 2 52 72 2 5 7 cos 75i s 55 88 " x s 7 475. sin 75i 5 sin 75i sin s " sin s s 0 6461, so s sin 1 0 6461 s 40 2i .

Therefore, by the Law of Sines, 5 7 475 7 475 50 72 . A triangle 1 r r sin 1 102 sin 72i . Thus, the area of the 19. (a) A sector 12 r 2 12 102 72 180 180 2 s 2 r 72

i shaded region is A shaded A sector A triangle 50 180 sin 72 s 15 3 m2 .

(b) The shaded region is bounded by two pieces: one piece is part of the triangle, the other is part of the circle.

S T The first part has length l 102 102 2 10 10 cos 72i 10 2 2 cos 72i . The second has length 4 . Thus, the perimeter of the shaded region is p l s 10T2 2 cos 72i 4 s 24 3 m. s 10 72 180 20. (a) If is the angle opposite the longest side, then by the Law of Cosines cos

92 132 202 0 6410. Therefore, 2 9 20

cos 1 0 6410 s 129 9i . (b) From part (a), s 129 9i , so the area of the triangle is A 12 9 13 sin 129 9i s 44 9 units2 . Another way to find T 9 13 20 a b c

21. Thus, the area is to use Heron’s Formula: A s s a s b s c , where s 2 2 T T A 21 21 20 21 13 21 9 2016 s 44 9 units2 .

21. Label the figure as shown. Now / 85i 75i 10i , so by the Law of Sines, 30 x

sin 75i sin 10i

%

x 30

sin 75i h . Now sin 85i sin 10i x

sin 75i sin 85i s 166 . h x sin 85i 3 0 sin 10i

z

%

Y

c

I

c

FOCUS ON MODELING Surveying 1. Let x be the distance between the church and City Hall. To apply the Law of Sines to the triangle with vertices at City Hall, the church, and the first bridge, we first need the measure of the angle at the first bridge, which is 180i 25i 30i 125i . x 0 86 0 86 sin 125i Then

% x

s 1 4089. So the distance between the church and City Hall is about sin 125i sin 30i sin 30i 1 41 kilometers.

510

Focus on Modeling

2. To find the distance z between the fire hall and the school, we use the distance found in the text between the bank and the cliff. To find z we first need to find the length of the edges labeled x and y.

bank

c

In the bank-second bridge-cliff triangle, the third angle is

1 55 sin 50i 180i 50i 60i 70i , so x s 1 26. sin 70i

c Y c

In the second bridge-school-cliff triangle, the third angle is

cliff

1 26 sin 45i s 1 09. sin 55i Finally, in the school-fire hall-cliff triangle, the third angle is 180i 80i 55i 45i , so y

180i 45i 80i 55i , so z

1 09 sin 80i

s 1 31. sin 55i Thus, the fire hall and the school are about 1 31 kilometers apart.

c

c Z c

c

fire hall

second bridge

c

c [

school

3. First notice that / D BC 180i 20i 95i 65i and / D AC 180i 60i 45i 75i .

AC 20 20 sin 45i From -AC D we get

% AC s 14 6i . From -BC D we get i i sin 45 sin 75 sin 75i 20 20 sin 95i BC

% BC

s 22 0. By applying the Law of Cosines to -ABC we get sin 95i sin 65i sin 65i T AB 2 AC 2 BC 2 2 AC BC cos 40i s 14 62 22 02 2 14 6 22 0 cos 40i s 205, so AB s 205 s 14 3 m.

Therefore, the distance between A and B is approximately 14 3 m.

4. Let h represent the height in meters of the cliff, and d the horizontal distance to the cliff. The third horizontal 200 sin 51 6i h angle is 180i 69 4i 51 6i 59i and so d % s 182 857. Then tan 33 1i i sin 59 d i i h d tan 33 1 s 182 857 tan 33 1 s 119 2 m.

5. (a) In -ABC, / B 180i , so / C 180i 180i . By the Law of Sines,

AB BC

sin sin

d sin sin

. sin sin d sin h h (b) From part (a) we know that BC . But sin . Therefore, % BC BC sin sin d sin h d sin sin BC

"h . sin sin sin 800 sin 25i sin 29i d sin sin (c) h

s 2350 m sin sin 4i " BC AB

6. Let the surveyor be at point A, the first landmark (with angle of depression 42i ) be at point B, and the other landmark 2430 2430 2430 s 3631 6. Similarly, sin 39i be at point C. We want to find BC . Now sin 42i % AB AB AC sin 42i 2430 s 3861 3. Therefore, by the Law of Cosines, BC 2 AB 2 AC 2 2 AB AC cos 68i " % AC sin 39i S BC 3631 62 3861 32 2 3631 6 3861 3 cos 68i s 4194. Thus, the two landmarks are approximately 4194 m apart.

Surveying

511

7. We start by labeling the edges and calculating the remaining angles, as shown in the first figure. Using the Law of Sines, 150 150 sin 29i b 150 150 sin 91i a

% a

s 83 97,

% b

s 173 18, we find the following: sin 29i sin 60i sin 60i sin 91i sin 60i sin 60i i i c 173 18 173 18 sin 32 d 173 18 173 18 sin 61

%c s 91 90,

%e s 151 67, sin 32i sin 87i sin 87i sin 61i sin 87i sin 87i i i 151 67 151 67 sin 41 f 151 67 151 67 sin 88 e

%e s 128 04,

% f s 195 04, sin 41i sin 51i sin 51i sin 88i sin 51i sin 51i 195 04 195 04 sin 50i h 195 04 195 04 sin 38i g

%g s 149 50, and

%h s 120 15. Note that i i i i i sin 50 sin 92 sin 92 sin 38 sin 92 sin 92i we used two decimal places throughout our calculations. Our results are shown (to one decimal place) in the second figure.

D

c

c

F c c

c

c E

C

H

G

c c c

c B

c

c I

8. Answers will vary. Measurements from the Great Trigonometric Survey were used to calculate the height of Mount Everest to be exactly 8839 m , but in order to make it clear that the figure was considered accurate to within a meter , the height was published as 8839.8 m . The accepted figure today is 88 48 m .