Basic Technical Mathematics, 9EAllyn J Washington Solution Manual

Type:

Solution Manual

Resource:

Basic Technical Mathematics

Edition:

9th Edition

Author(s):

Allyn J. Washington

Chapter 4

THE TRIGONOMETRIC FUNCTIONS 7. 50° , − 360° , − 30°

4.1

y

Angles

1. 145.6° + 2 ( 360° ) = 865.6° ⎛ 1° ⎞ ° 2. 17° 35′ = 17° 35′ ⎜ ⎟ = 17.58 ′ 60 ⎝ ⎠

x

8. 45° , 245° , − 250° y

°

3. −135

y

x x -135

o

9. positive: 45° + 360° = 405° negative: 45° − 360° = −315°

4. 240° y

10. 73° + 360° = 433° 240

o

73° − 360° = −287° x

11. −150° + 360° = 210° −150° − 360° = −510°

12. 462° + 360° = 822°

5.

462° − 2 ( 360° ) = −258°

y o

13. positive: 70° 30′ + 360 = 430° 30′

x

negative: 70° 30′ − 360° = −289° 30′

6. 330° , − 150° , 450° y

153° 47′ = −360° = −206°13′

15. 278.1° + 360° = 638.1°

o

x

96

14. 153° 47′ + 360° = 513° 47′

278.1° − 360° = −81.9°

Section 4.1 Angles

97

16. −197.6° + 360° = 162.4°

25. 56.0° = 0.977 rad to three significant digits

−197.6° − 360° = −557.6°

17. To change 0.265 rad to degrees multiply by

180

π

,

⎛ 180° ⎞ ° 0.265 rad ⎜ ⎟ ≈ 15.18 π rad ⎝ ⎠

26. 137.4° = 2.40 rad

18. To change 0.838 rad to degrees multiply by 180° / π , which gives 48.01; 0.0838 rad = 82.91°

19. To change 1.447 rad to degrees multiply by 180° / π , which gives 82.91; 1.447 rad = 82.91°

20. To change − 3.642 rad to degrees multiply by

27. 384.8° = 6.72 rad

180° / π , which gives − 208.67; − 3.642 rad = −208.67°

21. 0.329 rad ≈ 18.85°

28. −17.5° = −0.305 rad

22. 2.089 rad = 119.69°

29. 47° + 0.5° ⋅ 23. −4.110 rad = −235.49°

24. 6.705 rad = 384.17°

60′ = 47° + 30′ = 47°30′ 1°

30. 715.8° = 715° + 0.8° ⋅

60′ = 715° 48′ ° 1

31. −5.62° = −5° + 0.62° ⋅

60′ = −5°37′ 1°

32. 142.87° = 142° + 0.87° ⋅

33. 15°12′ = 15° + 12′ ⋅

60′ = 142° 52′ 1°

1° = 15.2° 60′

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

98

34. 517° 39′ = 517° + 39′ ⋅

35. 301°16′ = 301° + 16′ ⋅

36. −4° 47′ = −4° + 47′ ⋅

1° = 517.65° 60′

38. ( −3, 8 )

through ( −7, 5 ) .

°

1 = 301.27° 60′

1° = −4.78° 60′

37. Angle in standard position terminal side passing through ( 4, 2 ) .

41. Angle in standard position terminal side passing

42. ( −4, − 2 )

43. ( −2, 0 ) y

O

x (-2,0)

39. ( −3, − 5 )

44. ( 0, 6 ) y (0,6)

O

x

40. ( 6, − 1) 45. 31° , QI ; 310° , QIV 46. 180° is quadrantal; 92° , ° II 47. 435° , QI ; − 270° is quadrantal 48. −5° , QIV ; 265° , QIII

98

Section 4.2 Defining the Trigonometric Functions

49. 1 rad, QI; 2 rad, QII 50. 3 rad, QII ; π rad is quadrantal

51. 4 rad, QIII ;

π 3

rad, QI

52. 12 rad, QIV ; − 2 rad, QIII 53. 21° 42′36′′ = 21° + 42′ ⋅

1° 1° 1° + 36′′ ⋅ ⋅ = 21.710° 60′ 60′′ 60′

54. 7°16′23′′ = 7°16

23 16.4′ = 7°16.4′ = 7° = 7.273° 60′ 60

55. 0.274 × 60′ = 16.44 0.44′ × 60′′ = 26′′ so 86.274° = 86° 6′26′′ 60′ = 1.14′ 1° 60′′ 0.14′ ⋅ = 8′′ 1′ so 257.019° = 257°1′ 8′′

56. 0.019° ⋅

4.2

Defining the Trigonometric Functions

1. r = 42 + 32 = 25 = 5 3 sin θ = 5 4 cos θ = 5 3 tan θ = 4 5 csc θ = 3 5 sec θ = 4 4 cot θ = 3

99

4 ⇒ x = 7 2 − 42 = 33 7 33 = 0.8207 cos θ = 7 4 = 0.6963 tan θ = 33 7 csc θ = = 1.750 4 7 sec θ = = 1.219 33

2. sin θ =

cot θ =

33 = 1.436 4

3. r = 62 + 82 = 36 + 64 = 100 = 10 y 8 4 sin θ = = = r 10 5 x 6 3 cos θ = = = r 10 5 y 8 4 tan θ = = = x 6 3 r 10 5 csc θ = = = y 8 4 r 10 5 = = x 6 3 x 6 3 cot θ = = = y 8 4

sec θ =

4. r = 52 + 122 = 13 y 12 sin θ = = r 13 x 5 cos θ = = r 13 y 12 tan θ = = x 5 r 13 csc θ = = y 12 r 13 = x 5 x 5 cot θ = = y 12

sec θ =

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

100

5. r = 152 + 82 = 289 = 17 y 8 sin θ = = r 17 x 15 cos θ = = r 17 y 8 tan θ = = x 15 r 17 csc θ = = y 8 r 17 = x 15 x 15 cot θ = = y 8

sec θ =

6. r = 2402 + 7 2 = 62,500 = 250

8. r = 3.22 + 6.02 = 6.8 y 6.0 15 sin θ = = = r 6.8 17 x 3.2 8 cos θ = = = r 6.8 17 y 6.0 15 tan θ = = = x 3.2 8 r 17 csc θ = = y 15 r 17 = x 8 x 8 cot θ = = y 15

sec θ =

9. r = 1 + 15 = 16 = 4

sin θ =

y 70 7 = = r 250 25 x 240 24 cos θ = = = r 250 25 7 y 70 tan θ = = = x 240 24 r 250 25 csc θ = = = 7 y 70

sin θ =

r 250 25 = = x 240 24 x 240 24 cot θ = = = 7 y 70

sec θ =

sec θ =

7. r = 0.092 + 6.42 = 0.16 y 0.40 5 sin θ = = = r 0.16 2 x 0.09 9 cos θ = = = r 0.16 16 y 0.40 40 tan θ = = = x 0.09 9 r 0.16 2 csc θ = = = y 0.40 5 r 0.16 16 = = x 0.09 9 x 0.09 9 cot θ = = = y 0.40 40

sec θ =

y 15 = 4 r x 1 cos θ = = r 4 y tan θ = = 15 x 4 r csc θ = = y 15

r =4 x 1 x cot θ = = y 15

10. r = 3 + 22 = 7 y 2 sin θ = = r 7 x 3 = r 7 y 2 tan θ = = x 3

cos θ =

csc θ =

r 7 = y 2

sec θ =

r 7 = x 3

cot θ =

x 3 = y 2

Section 4.2 Defining the Trigonometric Functions

11. r = 12 + 11 = 2 sin θ =

y 1 2 = = r 2 2

x 1 2 = = r 2 2 y 1 tan θ = = = 1 x 1 r 2 csc θ = = = 2 y 1 cos θ =

r 2 = = 2 x 1 x 1 cot θ = = = 1 y 1

sec θ =

12. r = 62 + 52 = 61 5 y sin θ = = r 61 6 x cos θ = = r 61 y 5 tan θ = = x 6 61 r csc θ = = y 5 r 61 = x 6 x 6 cot θ = = y 5

sec θ =

13. r = 502 + 202 = 2900 = 10 29 y 20 2 sin θ = = = r 10 29 29 x 50 5 cos θ = = = r 10 29 29 y 20 2 tan θ = = = x 50 5 r 10 29 29 = csc θ = = y 20 2 29 r 10 29 = = 50 5 x x 50 5 = cot θ = = y 20 2

sec θ =

101

2

5 ⎛1⎞ 14. r = 1 + ⎜ ⎟ = 2 ⎝2⎠ 1 y sin θ = = r 5 2 x cos θ = = r 5 y 1 tan θ = = x 2 r csc θ = = 5 y r 5 = x 2 x cot θ = = 2 y

sec θ =

15. r = 0.687 2 + 0.9432 = 1.17 y 0.943 = 0.808 sin θ = = r 1.17 x 0.687 = 0.589 cos θ = = r 1.17 y 0.943 = 1.37 tan θ = = x 0.687 r 1.17 = 1.24 csc θ = = y 0.943 r 1.17 = = 1.70 x 0.687 x 0.687 = 0.729 cot θ = = y 0.943

sec θ =

16. r = 37.652 + 21.87 2 = 43.54 y 21.87 = 0.5023 sin θ = = r 43.54 x 37.65 = 0.8647 cos θ = = r 43.54 y 21.87 = 0.5809 tan θ = = x 37.65 r 43.54 = 1.991 csc θ = = y 21.87 r 43.54 = = 1.156 x 37.65 x 37.65 = 1.722 cot θ = = y 21.87

sec θ =

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

102

12 ⇒ x = 12 and r = 13 with θ in QL 13 r 2 = x 2 + y 2 ⇒ 169 = 144 + y 2 ⇒ y 2 = 25 y=5 y 5 x 12 sin θ = = , cot θ = = . r 13 y 5

17. cos θ =

1 y = 2 r r 2 = x 2 + y 2 ; 22 = x 2 + 1; x 2 = 3; x = ± 3

18. sin θ =

x 3 for acute θ = r 2 r 2 csc θ = = = 2 for acute θ y 1 cos θ =

y x r 2 = x 2 + y 2 = 12 + 22 = 5; r = ± 5 y 2 sin θ = = for acute θ r 5

19. tan θ = 2 =

r 5 sec θ = = = 5 for acute θ x 1

20. sec θ =

5 r = x 2 2

r 2 = x 2 + y 2 ; 5 = y 2 + 22 ; y 2 = 1; y = ±1 y 1 tan θ = = for acute θ x 2 x 2 cos θ = = for acute θ r 5

21. sin θ = 0.750 ⇒ y = 0.750 and r = 1 with θ in QI. r 2 = x 2 + y 2 ⇒ 12 = x 2 + 0.7502 ⇒ x 2 = 0.4375 ⇒ x = 0.661 x 0.661 = 0.881. cot θ = = y 0.750 1 r csc θ = = = 1.333. y 0.750

22. cos θ = 0.0326 =

x y

r 2 = x 2 + y 2 ; 1 = 0.0326 + y 2 ; y 2 = 0.999 y = ±0.999 y 0.999 sin θ = = = 0.999 r 1 y 0.999 tan θ = = = 30.6 x 0.0326

23. cot θ = 0.254 =

x y

r 2 = x 2 + y 2 ; r 2 = 0.2542 + 1; r 2 = 1.06; r = ±1.03 x 0.254 cos θ = = = 0.247 r 1.03 y 1 tan θ = = = 3.94 r 0.254

24. csc θ = 1.20 =

r y

r 2 = x 2 + y 2 ; 1.202 = x 2 + 1; x 2 = 0.44; x = ±0.663 r 1.20 sec θ = = = 1.82 x 0.663 x 0.663 cos θ = = = 0.553 r 1.20

25. For ( 3, 4 ) , r = 5, y 4 y 4 = and tan θ = = . r 5 x 3 For ( 6, 8 ) , r = 10,

sin θ =

y 8 4 y 8 4 = = and tan θ = = = . r 10 5 x 6 3 For ( 4.5, 6 ) , r = 7.5,

sin θ =

sin θ =

y y 6 4 6 4 = = and tan θ = = = . r 7.5 5 x 4.5 3

26. ( 5, 12 ) ; x = 5, y = 12, r = 25 + 144 = 13 cos θ =

5 5 , cot θ = 13 12

(15, 36 ) ; x = 15, y = 36, r = cos θ =

225 + 1296 = 39

15 5 15 5 = , cot θ = = 39 13 36 12

Section 4.2 Defining the Trigonometric Functions

( 7.5, 18) ; x = 7.5, y = 18,

30. sin θ =

r = 56.25 + 324 = 19.5 cos θ =

103

x = r 2 − y 2 = 32 − 22 = 5

7.5 5 7.5 5 = , cot θ = = 19.5 13 18 12

sec θ =

27. ( 0.3, 0.1) , r = 0.32 + 0.12 = 0.1 tan θ =

0.1 1 = 0.3 3

sec θ =

0.1 10 90 = = 0.3 3 9

2

y = y ⇒ r =1 r x cos θ = = x = r 2 − y 2 = 1 − sin 2θ r

32. cos θ =

90 sec θ = 9

tan θ =

( 33, 11) , r = 33 + 11 = 1210 = 11 10 tan θ =

2

31. sin θ =

3 1 = 9 3

2

r y 3 2 , tan θ = = = x x 5 5

⎛ 3 ⎞ ⎛ 2 ⎞ sec 2 − tan 2 θ = ⎜ ⎟ −⎜ ⎟ =1 ⎝ 5⎠ ⎝ 5⎠

( 9, 3) , r = 92 + 32 = 90 tan θ =

y 2 = ⇒ y = 2, r = 3 r 3

2

11 1 = 33 3

33. tan θ =

11 10 10 90 sec θ = = = 33 3 9

x = x ⇒ r =1 r y = x

r 2 − x2 1 − x2 = x x

6 4 y = = x −2 x + 1 6 x + 6 = −8 6 x = −14

28. ( 40, 30 ) , r = 40 + 30 = 50 2

2

50 5 = 30 3 40 4 cos θ = = 50 5

x=−

csc θ =

34.

(8, 6 ) , r = 82 + 62 = 10 10 5 = 6 3 18 4 cos θ = = 10 5

7 3

x 5 = 7 2 35 x= 2

csc θ =

x r 35. cos θ = ; = sec θ r x

( 36, 27 ) , r = 362 + 27 2 = 45 45 5 csc θ = = 27 3 36 4 cos θ = = 45 5 2

⎛3⎞ ⎛ 4⎞ 29. sin 2 θ + cos 2 θ = ⎜ ⎟ + ⎜ ⎟ ⎝5⎠ ⎝ 5⎠ 9 16 = = 25 25 25 = =1 25

2

y x ; cos θ = r r y x y r yr y ÷ = × = = = tan θ r r r x rx x

36. sin θ =

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

104

4.3

Values of the Trigonometric Functions

1. sin θ = 0.3527

θ = 20.65°

2. csc 27.82° =

1 sin 27.82°

= ( sin 27.82° )

−1

csc 27.82° = 2.143

3. cot θ = 0.345 1 tan θ = cotθ 1 tan θ = 0.345

θ = tan −1

1 0.345

θ = 71.0°

5. Answers may vary in exercises 5, 6, 7, 8. One set of measurements gives x = 7.6 and y = 6.5.

6.5 = 0.65 sin 40° = 10 7.6 cos 40° = = 0.76 10 6.5 tan 40° = = 0.86 7.6 10 csc 40° = = 1.54 6.5 10 sec 40° = = 1.32 7.6 7.6 cot 40° = = 1.17 6.5 6. sin θ = 0.97 cos θ = 0.26 tan θ = 3.73

csc θ = 1.04 sec θ = 3.86

7. sin θ = 0.26 cos θ = 0.97 tan θ = 0.27

csc θ = 3.86 sec θ = 1.04 cot θ = 3.73

8. sin θ = 0.80 cos θ = 0.60 tan θ = 1.33

csc θ = 1.25 sec θ = 1.66 cot θ = 0.75

9. sin 22.4° = 0.381

4. Find tan θ if sec θ = 2.504 1 1 = 2.504 cosθ 1 cos θ = 2504 ⎛ ⎛ 1 ⎞⎞ tan θ = tan ⎜ cos −1 ⎜ ⎟⎟ ⎝ 2.504 ⎠ ⎠ ⎝

sec θ =

tan θ = 2.296

10. cos 72.5° = 0.301

cot θ = 0.27

Section 4.3 Values of the Trigonometric Functions

11. tan 57.6° = 1.58

12. sin 36.0° = 0.588

13. cos15.71° = 0.9626

14. tan 8.653° = 0.1522

15. sin 89° = 1.00

16. cos 0.7° = 1.00

105

17. cot 67.78° = 0.4085

18. csc 22.81° = 2.579

19. sec 50.4° = 1.57

20. cot 41.8° = 1.12

21. csc 0.49° = 116.9

22. sec 7.8° = 1.01

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

106

23. cot 85.96° = 0.07063

29. tan θ = 0.207

θ = 11.7°

24. csc 76.30° = 1.029 30. sin θ = 0.109

θ = 6.3°

25. cos θ = 0.3261

θ = 70.97° 31. cos θ = 0.65007

θ = 49.453°

26. tan θ = 2.470

θ = 67.97° 32. tan θ = 5.7706

θ = 80.169°

27. sin θ = 0.9114 θ = 65.70 33. csc θ = 1.245

θ = 53.44°

28. cos θ = 0.0427

θ = 87.6° 34. sec θ = 2.045

θ = 60.73°

Section 4.3 Values of the Trigonometric Functions 35. cot θ = 0.0606 θ = 86.530

107

40. cot θ = 0.1519

θ = 81.36°

36. csc θ = 1.002

θ = 86.38°

41.

sin 43.7° = tan 43.7° cos 43.7° = 0.96

37. sec θ = 3.65

θ = 74.1°

42. sin 2 77.5° + cos 2 77.5° = 1

38. cot θ = 14.4

θ = 3.97° 43. tan 70° =

tan 30° + tan 40°

1 − tan 30° ( tan 40° )

= 2.747477419

39. csc θ = 8.26

θ = 6.95°

44. sin 78.4° = 2 ( sin 39.2° )( cos 39.2° ) = 0.9795752496

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

108

54.

45. y is always less than r.

y = 3x

y

46. For any acute angle θ , tanθ can equal any positive real number because it is the ratio of two positive numbers which can take on any positive value.

y = 2x

(x , 3 x) (x , 2 x)

θ2

θ1 x

x and for a given r , x decreases as θ r increases from 0° to 90° .

47. cos θ =

3x 2x − tan −1 x x −1 −1 θ1 − θ 2 = tan 3 − tan 2

θ1 − θ 2 = tan −1 48. sec θ =

r ≥ 1 since r ≥ x. x

θ1 − θ 2 = 8.1°

49. Given tanθ = 1.936 sin θ = sin ( tan −1 1.936 )

55. d = 70.0 + 30 cos θ

sin θ = 0.8885

d = 70.0 + 30 cos 54.4° d = 87 dB

50. sin θ = 0.6725 cos θ = cos ( sin −1 0.6725 )

56. l = a ( sec θ + csc θ )

cos θ = 0.7401

= 28.0 ( sec 34.5° + csc 34.5° ) = 83.4 cm

51. sec θ = 1.3698

(

tan θ = tan cos −1 (1.3698−1 ) tan θ = 0.93614

)

1.5λ 1.5 ( 200 ) = d 400 1.5 ( 200 ) θ = sin −1 400 θ = 48.6°

57. sin θ =

52. cos θ = 0.1063 csc θ =

1

sin ( cos −1 ( 0.1063) )

csc θ = 1.0057

58.

v = h tan θ 1.52 = 2.35 tan θ

53. y = 3x

y

(x , 3 x )

θ x

3x x −1 θ = tan 3

θ = tan −1

θ = 71.7°

tan θ =

1.52 ⎛ 1.52 ⎞ ° ; θ = tan −1 ⎜ ⎟ = 32.9 2.35 ⎝ 2.35 ⎠

Section 4.4 The Right Triangle

4.4

The Right Triangle

109

3.

1.

a = 79.552 − 65.822 = 44.68 65.82 cos A = 79.55 65.82 = 34.17° A = cos −1 79.55 B = 90° − 34.17° B = 55.83°

4. sin A = cos A =

7 65 4 65

= 0.868 = 0.496

7 = 1.75 4 4 sin B = = 0.496 65 4 = 0.868 cos B = 65 4 tan B = = 0.571 7 tan A =

2.

B = 90° − A a b a = b tan A b cos A = c b c= cos A tan A =

5. A 60° angle between sides of 3 in. and 6 in. determines the unique triangle shown in the figure below.

3 in.

60

a ⇒ a = b tan A b a = 7.60 tan 50° a = 9.06 b 7.60 cos A = ⇒ c = = 11.8 c cos 50° B = 90° − 50°

o

6 in.

tan A =

B = 40

3.

6. Once the given parts are in place two sides must be included. In order to make the angles 40° and 50° respectively, only two lengths and two directions of the lines will close up the space to form a triangle.

°

40

o

50 4 in.

o

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

110

7. Once the given parts are in place one side remains to be included. Only one length and one direction of the line will close up the space to form a triangle.

12. b = 12402 − 9322 = 818 a 932 sin A = = c 1240 ⎛ 932 ⎞ ° A = sin −1 ⎜ ⎟ = 48.7 1240 ⎝ ⎠ B = 90° − A B = 41.3°

8. Once given parts are in place only two lengths and two directions of the lines will close up this space to form a triangle.

6700 ⇒ c = 6850 c ∠B = 90° − 77.8° = 12.2°

9. sin 77.8° =

tan 77.8° =

6700 ⇒ b = 1450 b

10. B = 90.0° − 18.4° = 71.6° a sin A = c a = c sin A = 0.0897 ( sin 18.4° ) = 0.0283 b c b = c ( cos A ) = 0.0897 ( cos 18.4° ) = 0.0851

cos A =

11. b = 3452 − 1502 = 311 a 150 sin A = = c 345 ⎛ 150 ⎞ ° A = sin −1 ⎜ ⎟ = 25.8 ⎝ 345 ⎠ a 150 cos B = = = 0.435 c 345 ⎛ 150 ⎞ ° B = cos −1 ⎜ ⎟ = 64.2 ⎝ 345 ⎠

13. ∠A = 90° − 32.1° = 57.9° , b sin 32.1° = ⇒ b = 12.6 23.8 a cos 32.1° = ⇒ a = 20.2 23.8 14. A = 90° − 64.3° = 25.7° b sin B = c b 0.652 c= ;c= = 0.724 sin B sin 64.3° b tan B = a b 0.652 a= = = 0.314 tan B tan 64.3° 15. a = 882 − 822 = 32 B 82 cos A = = c 88 ⎛ 82 ⎞ A = cos −1 ⎜ ⎟ = 21° ⎝ 88 ⎠ B 82 sin B = = c 88 ⎛ 82 ⎞ B = sin −1 ⎜ ⎟ = 69° ⎝ 88 ⎠ 16. c = 59202 + 41102 = 7210 a 5920 tan A = = c 4110 ⎛ 5920 ⎞ ° A = tan −1 ⎜ ⎟ = 55.2 ⎝ 4110 ⎠ b 4110 tan B = = z 5920 ⎛ 4110 ⎞ ° B = tan −1 ⎜ ⎟ = 34.8 5920 ⎝ ⎠

Section 4.4 The Right Triangle

17. ∠B = 90° − 32.1° = 57.9° , a sin 32.1° = ⇒ a = 30.21 56.85 b cos 32.1° = ⇒ b = 48.16 56.85 18. A = 90.0° − 12.60° = 77.40° b sin B = c b = c sin B = 18.42 sin 12.60° = 4.018 a cos B = c a = c cos B = 18.42 cos 12.60° = 17.98 19. c = 56.732 + 44.092 = 71.85 a 56.73 tan A = = b 44.09 ⎛ 56.73 ⎞ ° A = tan −1 ⎜ ⎟ = 52.15 ⎝ 44.09 ⎠ b 44.09 tan B = = a 56.73 ⎛ 44.09 ⎞ ° B = tan −1 ⎜ ⎟ = 37.85 ⎝ 56.73 ⎠ 20. b = 12.632 − 9.9082 = 7.833 a 9.908 sin A = = c 12.63 ⎛ 9.908 ⎞ ° A = sin −1 ⎜ ⎟ = 51.67 ⎝ 12.63 ⎠ a 9.908 cos B = = c 12.63 ⎛ 9.908 ⎞ ° B = cos −1 ⎜ ⎟ = 38.33 12.63 ⎝ ⎠ 21. ∠A = 90° − 37.5° = 52.5° , b tan 37.5° = ⇒ b = 0.661 0.862 0.862 cos 37.5° = ⇒ c = 1.09 c

111

22. B = 90° − 87° = 3° b cos A = c b 8.450 c= = = 176.1 cos A cos 87.25° a tan A = b a = b tan A = 8.450 tan 57.25° = 175.9 23. A = 90.0° − 77.18° = 15.82° b sin B = c b 1.849 c= = = 1.922 sin B sin 74.18° b tan B = a b 1.849 a= = = 0.5239 tan B tan 74.18° 24. B = 90.0° − 51.36° = 38.64° a sin A = c 3692 a = = 4727 c= sin A sin 51.36° a tan A = b 3692 a = = 2952 b= tan A tan 51.36° 591.87 ⇒ A = 65.89° ; 264.93 264.93 ⇒ B = 24.11° tan B = 591.87

25. tan A =

c = 264.932 + 591.87 2 = 648.46

26. a = c 2 − b 2 = 5.08642 − 2.9507 2 = 4.1430 b 2.9507 sin B = = c 5.0864 ⎛ 2.9507 ⎞ ° B = sin −1 ⎜ ⎟ = 35.459 ⎝ 5.0864 ⎠ b 2.9507 cos A = = c 5.0864 ⎛ 2.9507 ⎞ ° A = cos −1 ⎜ ⎟ = 54.541 ⎝ 5.0864 ⎠

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

112

27. B = 90° − 2.975° = 87.025° b cos A = c 14.592 b c= = = 14.611 cos A cos 2.975° a tan A = b a = b tan A = 14.592 tan 2.975° = 0.7585 28. A = 90.0° − 84.942° = 5.058° a cos B = c a 7413.5 c= = = 84, 087 cos B cos 84.942° b tan B = a b = a tan B = 7413.5 tan 84.942° = 83, 760 29. A = 90.0° − 9.56° = 70.44° b sin B = c b 1.849 = = 1.922 c= sin B sin 74.18° b tan B = a b sin 9.56° = 0.00973 b = 0.00162 a cos B = c a cos 9.56° = 0.00973 a = 0.00959 1.28 16.3 A = 4.9°

30. tan A =

16.3 1.28 B = 85.5°

32. The given information does not determinbe a unique triangle.

33. sin 61.7° =

C = 16.4

31. The given information does not determinbe a unique triangle.

19.7 36.3

34. tan A =

⎛ 19.7 ⎞ ° A = tan −1 ⎜ ⎟ = 28.5 ⎝ 36.3 ⎠ 0.6673 0.8742 ⎛ 0.6673 ⎞ ° A = cos −1 ⎜ ⎟ = 40.24 ⎝ 0.8742 ⎠

35. cos A =

x 7265 x = 7265 sin 22.45°

36. sin 22.45° =

= 2774

37.

B 37.5 25.6 C

A

b

25.6 37.5 A = 43.1° < B = 90° − 43.1° = 47.9°

sin A =

38.

B c

tan B =

C = 1.282 + 16.32

3.92 3.92 ⇒x= = 4.45 x sin 61.7°

52.3

o

a C

A 8.50

8.50 a 8.50 a= tan 52.3° a = 6.57

tan 52.3° =

Section 4.5 Applications of Right Triangles

113

4.5

39.

B 827 17.6

A

a

o

Applications of Right Triangles

1. 90° − 62.1° = 27.9° d = 2090 ft

C

b

b 827 b = 827 cos17.6°

cos17.6° =

2.

317

b = 788

.8 k

26.55

m

o

h

d = rt

40.

⎛ 1 ⎞ d = ( 6355 km/h )( 3.000 min ) ⎜ ⎟ ⎝ 60 min ⎠ d = 317.8 km h 317.8 h = 317.8sin 26.55°

sin26.55° =

0.842 tan B = 0.596 B = tan −1

h = 142 km

B = 54.7°

41. ∠B = 90° − ∠A, a A = ⇒ a = c sin A, c b cos A = ⇒ b = c cos A c b a a tan A = b

42. tan B =

3. sin 54° =

h ⇒ h = 120 sin 54° = 97 ft 120

4.

b a −1 a A = tan b

B = tan −1

c = a 2 + b2 4.25 4.25 ⇒ θ = cos −1 50.5 50.5 4.25 − x ° cos (θ + 0.5 ) = 50.5 4.25 ⎛ ⎞ + 0.5° ⎟ x = 4.25 − 50.5cos ⎜ cos −1 50.5 ⎝ ⎠ x = 0.44 m cos θ =

a c b tan B = a A = 90.0° − B

43. cos B =

c=

a cos B

b = a tan B

44. Yes, angles A and B. At least one side must be given in order to find a unique solution of the triangle.

h 22.8 h = 22.8 tan 62.6° = 44.0 ft

5. tan 62.6° =

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

114

h 1.25 h = 1.25 sin 13.0° = 0.281 m

7. 25.0 ft ⋅

θ = sin −1

15. tan A =

4.63 12.60 ⎛ 4.63 ⎞ ° A = tan −1 ⎜ ⎟ = 20.2 ⎝ 12.60 ⎠

8. tan A =

196.0 h 196.0 = 850.1 cm h= cos 76.67°

16. Diameter = 12.3 m, each section is

h 10. tan 16° = 5200 h = 5200 tan 16° h = 1500 ft

x 6.15 x = 6.15 sin 6° = 0.6429 2 x = 1.29 m

sin 6° =

17. θ = tan −1

325 12. tan 7.5 = h 325 = 2470 ft h= tan 7.5° °

18.

13.

h 9.8 h = 9.8 tan1.2° h = 0.21 mi

tan1.2° =

h

6.0 = 3.4° 100

360° 45° = 45° , so = 22.5° 8 sides 2 0.375 sin 22.5 = x 0.375 x= = 0.9799 sin 22.5° 2 x = 1.96 m

19. tan α =

6.75 ⎛ 6.75 ⎞ ° ; α = tan −1 ⎜ ⎟ = 23.4 15.5 15.5 ⎝ ⎠

h 12.5 h = 12.5 sin 20.0° = 4.3 ft Height of the light above the street:

20. sin 20.0° = 14.

360° = 12°.

30

so r = 6.15 m

2.65 2.65 ⇒x= = 25.4 ft x sin 6.0°

o

8.00 4.00

⎛ 8.00 ⎞ ° A = tan −1 ⎜ ⎟ = 63.4 4.00 ⎝ ⎠ 4.00 tan B = 8.00 ⎛ 4.00 ⎞ ° B = tan −1 ⎜ ⎟ = 26.6 ⎝ 8.00 ⎠ The angles are 26.6° , 63.4° and 90.0°

9. cos 76.67° =

1.2 9.8

3.50 5.00

θ = 44.4°

12 in.

2.00 = 300 in., θ = tan −1 = 0.4° 1 ft 300

11. sin 6.0° =

3.50 5.00

sin θ =

6. sin 13.0° =

4.3 + 28.0 = 32.3 ft

21. θ = sin −1

12.0 = 8.1° 85.0

Section 4.5 Applications of Right Triangles

22. cos A =

115

27. These are similar triangles.

9.3 x = x + 49 22.5 22.5 x = 9.3 ( x + 49 )

76 − 8 2 76 2

A = 26.5°

22.5 x = 9.3 x + 455.7 13.2 x = 455.7

a 23. cos 78° = 3.65 a = 3.65 cos 78° = 0.7589 cm

x = 34.5 mm tan

x = 3.83 − 0.7589 = 3.07 cm (North)

θ 2

θ 24. 90° − 39.7° = 50.3° R sin 50.3° = R + 552 ( R + 552 ) sin 50.3° = R

2

=

22.5 = 0.2695 49 + 34.5

= 15.08°

θ = 30.2° 28. C = 2π ⋅ 3960 ⋅ cos 66° 32′ C = 9910 mi

0.7694 R + 424.7 = R 424.7 = 0.2306 R R = 1840

29.

32.0

°

360 = 72°. Radii 5 drawn from the center of the pentagon (which is also the center of the circle) through adjacent vertices of the pentagon outward to the fence form an isosceles triangle with base 92.5 and equal sides x. A ⊥ bisector from the center of the pentagon to the base this isosceles triangle forms a right triangle with hypotenuse x and base 46.25. The

o

25. Each angle of the pentagon is

4.50 in.

x

x 4.50 x = 1.24

sin/ 6.0° =

180° − 72° base angle of this right triangle is = 54°.

2 Thus, 46.25 46.25 cos ( 54° ) = ⇒x= x cos 54° and C = 2π ( x + 25 )

distance between ends = 2 x = 2.48 in.

30.

⎛ 46.25 ⎞ C = 2π ⎜ + 25 ⎟ = 651 ft ° ⎝ cos54 ⎠

H

θ

x 26. Find x, tan82.18° = 482.6 x = 482.6 tan 82.18° = 3514 ft Find h, tan 3.6° =

h 3514

W

W 16 H 9 = ⇒ = H W 16 9 H 9 tan θ = = W 16 θ = 29°

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

116

31.

32.

1 1 C = ⎡⎣ 2π (11.8 + 1) ⎤⎦ = 12.8π 2 2 12.8π ° cos 31.8 = ⇒ l = 47.3 m l x tan 9.9 = 7800 x = 7800 tan 9.9°

36.

A = bh h sin 85° = ⇒ h = 48 sin 85° 48 A = bh = 92 ( 48 sin 85° ) A = 4400 cm 2

°

37.

y tan 12.1 = 7800 y = 7800 tan 12.1° Antenna = y − x °

= 7800 tan 12.1° − 7800 tan 9.9° = 310 ft

33.

h ⇒ h = a sin θ a x cos θ = ⇒ x = a cos θ a b + (b + 2x ) A= ⋅h 2 2b + 2a cos θ = ⋅ a sin θ 2 A = ( b + a cos θ ) ⋅ a sin θ sin θ =

5380 6680 ⎛ 5380 ⎞ ° D = sin −1 ⎜ ⎟ = 53.7 ⎝ 6680 ⎠ 2860 tan 53.7° = h 2860 = 2100 ft h= tan 53.7° sin D =

38. r 34. cos 27.8° = 7.25 r = 7.25 cos 27.8° = 6.41 7 sin 27.8° = 7.25 t = 7.25 sin 27.8° = 3.38 m 5.25 − 3.38 = 1.87 m m tan 39.4° = 1.87 m = 1.87 tan 39.40° = 1.54 m Therefore, x = s + r = 6.41 + 1.54 = 7.95 m

35. tan

θ

d

= 2 2 x =

6400 6400 + 35,300 6400 θ = cos −1 6400 + 35,300

cos θ =

θ = 81.2° 39. Let e = length of edge of cube, then a diagonal in the base has length 2e. If θ = angle between the base and a diagonal of the cube, then e tan θ = 2e θ = 35°

d 2

tan θ2

= 0.5d cot

θ 2

40. tan

θ 2

θ 2

=

42.5 2

x

= 3.2°

θ = 6.4°

Chapter 4 Review Exercises

117

Chapter 4 Review Exercises 1. 17.0° + 360.0° = 377.0° , 17.0° − 360.0° = −343.0° 2. 248.3° + 360° = 608.3° , 248.3° − 360° = −111.7° 3. −217.5° + 360° = 142.5° , − 217.5° − 360° = −577.5° 4. −7.6° + 360° = 352.4° , − 7.6° − 360° = −367.6° ⎛ 1° ⎞ ° 5. 31° + 54′ ⎜ ⎟ = 31.9 ′ 60 ⎝ ⎠ ⎛ 1° ⎞ ° 6. 574° 45′ = 574° + 45′ ⎜ ⎟ = 574.75 ⎝ 60′ ⎠ ⎛ ⎛ 1 ⎞⎞ ° 7. −38° 6′ = − ⎜⎜ 38° + 6′ ⎜ ⎟ ⎟⎟ = −38.1 ′ 60 ⎝ ⎠⎠ ⎝ °

⎛ 1° ⎞ ° 8. 321° 27′ = 321° + 27′ ⎜ ⎟ = 321.45 ′ 60 ⎝ ⎠ ⎛ 60′ ⎞ 9. 17.5° = 17° + 0.5° ⎜ ° ⎟ = 17° 30′ ⎝1 ⎠

⎛ ⎛ 60′ ⎞ ⎞ 10. −65.4° = − ⎜ 65° + 0.4° ⎜ ° ⎟ ⎟ = −65° 24′ ⎝ 1 ⎠⎠ ⎝ ⎛ 60′ ⎞ 11. 749.7° = 749° + 0.7° ⎜ ° ⎟ = 749° 42′ ⎝1 ⎠

⎛ 60′ ⎞ 12. 126.25° = 126° + 0.25° ⎜ ° ⎟ = 126°15′ ⎝1 ⎠

14. r 2 = x 2 + y 2 = 52 + 42 = 41 sin θ =

4

cos θ =

5

41

41 4 tan θ = , 5

y 7 = , r 25 x 24 cos θ = = , r 25 y 7 tan θ = = , x 24

r 25 = y 7 r 25 sec θ = = x 24 x 24 cot θ = = y 7

csc θ =

csc θ =

41 4

,

sec θ =

41 5

cot θ =

5 4

15. r 2 = x 2 + y 2 = 482 + 482 ⇒ r = 48 2 48 1 sin θ = = , csc θ = 2 48 2 2 48 1 cos θ = = , sec θ = 2 48 2 2 48 tan θ = = 1, tan θ = 1 48 16. r 2 = x 2 + y 2 = 1.22 + 0.52 ⇒ r = 1.3 0.5 5 13 = , sin θ = csc θ = 1.3 13 5 1.2 12 13 = , cos θ = sec θ = 1.3 13 12 0.5 5 12 = , tan θ = cot θ = 1.2 12 5 17. r 2 = x 2 + y 2 ⇒ 132 = x 2 + 52 ⇒ x = 12 x 12 cos θ = = = 0.923 r 13 x 12 cot θ = = = 2.40 y 5 18. r 2 = x 2 + y 2 ⇒ 82 = 32 + y 2 ⇒ y = 55 sin θ =

y 55 = = 0.927 r 8

tan θ =

y 55 = = 2.47 x 3

13. r = x 2 + y 2 = 242 + 7 2 = 625 = 25 sin θ =

,

19. tan θ = 2 =

y ⇒ y = 2 x, r 2 = x 2 + y 2 = 5 x 2 ⇒ r x = 5x

cos θ =

x x = = 0.447 r 5x

csc θ =

r 5x = = 1.12 y 2x

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

118

20. cot θ =

x = 40 ⇒ x = 40 y y

r 2 = x 2 + y 2 ⇒ r 2 = 1600 y 2 + y 2 ⇒ r = 1601y sin θ =

y y = = 0.0250 r 1601y

r 1601 y = 1.0003 sec θ = = x 40 y

21. sin 72.1° = 0.952 from the calculator 22. cos 40.3° = 0.763 from the calculator 23. tan 88.64° = 31.12 from the calculator 24. sin 0.91° = 0.01588 from the calculator 25. sec 18.4° = 1.05 from the calculator 26. csc 82.4° =

1 = 1.01 from the calculator sin 82.4°

27. ( cot 7.06° )( sin 7.06° ) − cos 7.06° sin 7.06° − cos 7.06° tan 7.06° = 0 from the calculator =

28. ( sec 79.36° )( sin 79.36° ) − tan 79.36° °

sin 79.36 − tan 79.36° ° cos 79.36 = 0 from the calculator =

29. cos θ = 0.950 ⇒ θ = cos −1 ( 0.950 )

θ = 18.2° from the calculator 30. sin θ = 0.63052 ⇒ θ = sin −1 ( 0.63052 ) = 39.088° from the calculator

31. tan θ = 1.574 ⇒ θ = tan −1 (1.574 ) = 57.57° from the calculator

32. cos θ = 0.0135 ⇒ θ = cos −1 ( 0.0135 ) = 89.23° from the calculator

33. csc θ = 4.713 ⇒

θ = sin −1

1 1 = 4.13 ⇒ sin θ = sin θ 4.713

1 = 12.25° from the calculator 4.713

1 1 = 0.7561 ⇒ tan θ = tan θ 0.7561 1 θ = tan −1 = 52.91° from the calculator 0.7561

34. cot θ =

1 1 = 25.4 ⇒ cos θ = cos θ 25.4 1 θ = cos −1 = 87.7° from the calculator 25.4

35. sec θ =

1 1 = 1.92 ⇒ sin θ = sin θ 1.92 1 θ = sin −1 = 31.4° from the calculator 1.92

36. csc θ =

1 = 7.117 tan θ 1 tan θ = 7.117 1 θ = tan −1 7.117 ° θ = 8.00 from the calculator

37. cot θ =

1 1 = 1.006 ⇒ cos θ = cos θ 1.006 1 θ = cos −1 = 6.26° from the calculator 1.006

38. sec θ =

39. sin θ = 0.9998

θ = sin −1 0.9998 = 88.85° from the calculator 40. There is no θ for which cos θ = 1.402.

41. ∠B = 90.0° − 17.0° = 73.0° a tan 17.0° = ⇒ a = 1.83 6.00 6.00 6.00 cos 17.0° = ⇒c= = 6.27 c cos 17.0°

Chapter 4 Review Exercises

119

42. ∠A = 90° − 68.1° = 21.9° b tan 68.1° = ⇒ b = 2690 1080 1080 cos 68.1° = ⇒ c = 2900 c

50. ∠A = 90° − B = 90° − 4.38° = 85.62° 5682 sin B = sin 4.38° = ⇒ c = 74, 400 c 5682 ⇒ a = 74,180 tab B = tan 4.38° = a

43. c 2 = a 2 + b 2 = 81.02 + 64.52 ⇒ c = 104 81.0 tan A = ⇒ A = 51.5° 64.5 B = 90° − A = 38.5°

51. c 2 = a 2 + b 2 ⇒ 15.4832 = 11.6522 + b 2 ⇒ b = 10.196 11.652 sin A = ⇒ A = 48.813° 15.483 B = 90° − A = 41.187°

44. c 2 = a 2 + b 2 = 3822 + 1062 ⇒ b = 367 106 tan A = ⇒ A = 16.1° 382 B = 90° − A = 73.9° 45. ∠B = 90.0° − 37.5° = 52.5° 12.0 tan 37.5° = ⇒ b = 15.6 b 12.0 12.0 ⇒c= = 19.7 sin 37.5° = c sin 37.5°

52. c 2 = a 2 + b 2 = 724.392 + 852.442 ⇒ c = 1118.7 724.39 tan A = ⇒ A = 40.357° 852.44 B = 90° − A = 49.643° 53. x 25

o

o

12 - d °

°

°

65 d

°

46. ∠A = 90 − B = 90 − 15.7 = 74.3 a cos B = cos 15.7° = ⇒ a = 121 126 b sin B = sin 15.7° = ⇒ b = 34.1 126 47. c 2 = a 2 + b 2 ⇒ 7.6422 = a 2 + 6.5082 ⇒ a = 4.006 6.508 cos A = ⇒ A = 31.61° 7.642 B = 90° − A = 58.39° 48. c 2 = a 2 + b 2 = 0.7212 + 0.1442 ⇒ c = 0.735 0.721 tan A = ⇒ A = 78.7° 0.144 B = 90° − A = 11.3° 49. ∠B = 90.0° − 49.67° = 40.33° a sin 49.67° = ⇒ a = 0.6292 0.8253 b cos 49.67° = ⇒ b = 0.5341 0.8253

90° − 25° = 65° x x ⇒d = tan 65° d x x tan 25° = = 12 − d 12 − tanx65° tan 65° =

x ⎞ ⎛ ° x = ⎜ 12 − ⎟ tan 25 − 12 d ⎝ ⎠ tan 25° x = 12 tan 25° − x ⋅ tan 65° ° ⎛ tan 25 ⎞ = 12 tan 25° x ⎜1 + ° ⎟ tan 65 ⎝ ⎠ x=

12 tan 25° 25° 1 + tan tan 65°

x = 4.6

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

120

54. (1) sin 31° =

58.

x ⇒ x = 1.03 2

(2) 90° − 31° = 59° , cos 59° =

x ⇒ x = 1.03 2

2.607 A

b ⇒ b = 2 cos 31° 2 22 = b 2 + x 2 = ( 2 cos 31° ) + x 2 ⇒ x = 1.03

(3) cos 31° =

4.517

2.607 ⇒ A = 29.99° 4.517

tan A =

Method (1) is the easiest

59.

55.

s

x ⇒ x = h cot A h y cot B = ⇒ y = h cot B h c = x + y = h cot A + h cot B cot A =

⇓ 45

o

10

10

60. sin s

s

2

2

c A = a sin 2 2 A c = 2a sin 2

Each side s of the octagon is the base of an isosceles triangle with vertex angle =

360° = 45°.

8

s

sin 22.5° = 2 ⇒ s = 20sin 22.5° 10 perimeter = 8s = 8 ( 20sin 22.5° ) = 61.2

A 2c = 2 a

61.

1+x2

x

θ

56. For a given r , y increases as θ increases from

tan θ = x =

1

0° to 90°.

csc θ =

57. y ( x, 7)

x

2 7 = 3 x 2 x = 21

x = 10.5

y (2, 3)

(3, 2)

tan θ =

62.

B A

(3, 2)

x

x 1

x2 + 1 x

Chapter 4 Review Exercises

2 ⇒ A = 33.7° 3 3 tan ( A + B ) = ⇒ A + B = 56.3° 2 ° B = 56.3 − A = 22.6° tan A =

121

P cos θ

68. S = P sec θ = S=

12.0 = 13.8 V ⋅ A cos 29.4°

69. (a) A triangle with angle θ included between sides a and b, the base, has an altitude of a sin θ . The

63.

7.50

1 ⋅ b ⋅ a sin θ .

2

4.50

area, A, is A =

7.50

(b) The area of the tract is A =

A

1 ⋅ 31.96 ⋅ 47.25 2

sin 64.09° = 679.2 m 2 .

15.0

70.

4.50 15.0 A = 16.7°

tan A =

b

h

θ

1 base ( base ) ⋅ h, cot θ = 2 h A = bh + h 2 cot θ

(a) A = bh + 2 ⋅

64.

2.8

(b) A = (12.6 )( 4.75 ) + 4.752 cot 37.2°

o

A = 89.6 ft 2

1.25 71. altitude = 27.5 sin 10.3° = 4.92 km d 2

72. θ = sin −1

d = distance between extreme positions ⎛ 5.6° ⎞ = 2 ⎜ 1.25sin ⎟ 2 ⎠ ⎝

8.5 = 32° 16.0

73. l

65. e = E cos α ⇒ 56.9 = 339 cos α ⇒ α 56.9 = cos −1 = 80.3° 339 1 d1d 2 sin θ 2 1 = ( 3.25 )( 4.38 ) sin 72.0° 2 = 6.77 ft 2

42.5

o

1.50

66. A =

67. θ = tan −1

v2 80.7 2 = tan −1 = 12.1° rg ( 32 )( 950 )

74. h = 85 sin 52° = 67 ft

A = lw

1.50 ( 4.80 ) cos 42.5° A = 9.77 ft 2 A=

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

122

75.

12.8

78.

z z x

θ = 17.38

y

R = 1750 ohms

48 o 28 o

R R 1750 ⇒z= = z cos θ cos17.38° z = 1830 Ω

cos θ =

w

76° − 26° = 48° 12.8 12.8 ⇒z= z tan 48° 12.8 12.8 sin 48° = ⇒x= x sin 48° y 12.8 sin 48° = ⇒ y = x sin 28° = sin 28° x sin 48° w 12.8 cos 28° = ⇒ w = x cos 28° = cos 28° x sin 48° tan 48° =

1 1 (12.8 ) z + wy 2 2 ° 1 12.8 1 ⎛ 12.8 ° ⎞ ⎛ 12.8sin 28 ⎞ A = (12.8 ) cos 28 + ⎜ ⎟ ⎜ ⎟ ° 2 tan 48° 2 ⎝ sin 48° ⎠ ⎝ sin 48 ⎠ A=

A = 135 ft 2

76.

75.0

h

50.0 65

o

79.

source

65 km

h

0.03

o

city h 65 h = 34 m

sin 0.03° =

80. tan 76° =

ceiling ⇒ ceiling = 3800 m 950

81. Let x = length of window through which sun does not shine. x + 2.5 tan 65° = ⇒ x = 2.0 tan 65° − 2.5 2.0 2.0 tan 65° − 2.5 = 56% percent window shaded = 3.2

o

x

x

82.

D

h sin 65.0 = ⇒ h = 50.0sin 65.0° 50.0 x ° cos 65.0 = ⇒ x = 50.0 cos 65.0° 50.0 °

d 2800 ft θ

25

o

D

x

1 1 V = ( b1 + b2 ) ⋅ h ⋅ l = ( 75.0 + ( 75.0 + 2 x ) ) h ⋅ l 2 2 1 V = 150 + 2 ( 50.0 cos 65.0° ) ( 50.0sin 65.0° ) ( 5280 ) 2 V = 23, 000, 000 ft 3

(

77. d = a + b =

)

1.85 1.85 + = 4.43 m ° tan 28.3 tan ( 90.0° − 28.3° )

d = 1130t ; sin 25° =

2800 2800 ⇒d = d sin 25°

d 2800 = 1130 1130sin 25° 2800 ( 660 ) D = 660t = 1130sin 25° 2800 2800 tan 25° = ⇒x= −D x+D tan 25° 2800 2800 tan θ = = 2800 ( 660 ) 2800 x ° − °

t=

tan 25

θ = 53

°

1130 sin 25

Chapter 4 Review Exercises

83. sin 31.0° =

123

d 14.2 sin 21.8° ⇒ sin 31.0° = ⇒x x x 14.2 sin 21.8° = = 10.2 in. sin 31.0°

88.

0.355 in.

t t x

t

x

84.

θ

beach

2.75 km 69.0

boat

o

sin

d

2.75 2.75 ⇒d = sin 69° d 2.75 60 min ⋅ d = 37.5t ⇒ t = ° 37.5sin 69.0 1h t = 4.71 min

θ

sin 69° =

sin

d = distance from helicopter to end of span 1160 tan 2.2° = ⇒ d = 30, 200 m d II. Line of sight perpendicular to middle of span o

1160 2

1160 2

d

⇒ θ = 58.2°

( 5 − 2 ) ⋅180

22.5 tan 54° . The area of 12 such pentagons is 1 12 ⋅ 5 ⋅ ⋅ 45.0 ⋅ 22.5 tan 54° or 41,807.60083 mm 2 .

2 ( 6 − 2 ) ⋅180 Each angle of a regular hexagon is 6 ° = 120 . A regular hexagon with a side of 45.0 mm of 22.5 tan 60° . The area of 20 such hexagons is

⇒ d = 30, 200 m

l ⇒l 2.25 = 2.28 m

86. length of side piece = l ⇒ sin 80.0° =

b = length of base = 2.25 + 2 x where tan 80.0 2.25 = x 1 A = ( 2.25 + 2 x + 2.25 )( 2.25 ) 2 1⎛ 2.25 ⎞ 2 A = ⎜ 4.50 + 2 ⎟ ( 2.25 ) = 5.96 m 2⎝ tan 80.0° ⎠

87. sin

−0.1802 / 2 0.180 − 0.355 2 2

consists of 6 triangles of base 45.0 and an altitude

d

tan1.1° =

0.180 / 2

consists of 5 triangles of base 45.0 and altitude of

d

1.1

2

=

= 5 108°. A regular pentagon iwth a side of 45.0 mm

1160

o

θ

t / 2 0.355 t / 2 0.355 = 2 ⇒ = 2 x x+t x x+t −t 2 / 2 x= t / 2 − 0.355 2

89. Each angle of a regular pentagon is

85. I. Line of sight perpendicular to end of span 2.2

2

=

0.00200° d /2 = ⇒ d = 1.83 km 2 52,500

°

1 20 ⋅ 6 ⋅ ⋅ 45 ⋅ 22.5 tan 60° = 105, 222.0866 mm 2 .

2 Thus, the area of 12 regular pentagons of side 45.0 mm and 20 regular hexagons of side 45.0 mm is 147,029.6874 mm 2 (147, 000 mm 2 rounded off ) .

Since this is the area of a flat surface it approximates the area of the spherical soccer ball which is given by 2

⎛ 222 ⎞ 2 4π r 2 = 4π ⋅ ⎜ ⎟ = 154,830.2523 mm ⎝ 2 ⎠

(155, 000 mm rounded off ) .

2

90. θ = tan −1

1.25 = 30.8° 2.10

Chapter 4 THE TRIGONOMETRIC FUNCTIONS

124

91.

tan

67.5° d = 2 5.0

67.5° 2 67.5° l = 5.0 + 65.0 + 5.0 tan = 73.3 cm 2 d = 5.0 tan

x 42.0

o

d

25.0

o

94.

y

o

375

30

x ⇒ x = d tan 42.0° d d cos 25.0° = ⇒ d = 375cos 25.0° 375 x = 375cos 25.0° tan 42.0° x sin 25.0° = ⇒ x = 375sin 25.0° 375 x + y = 375cos 25.0° tan 42.0° + 375sin 25.0° tan 42.0° =

x

w

x = 252 + 752 w/2 ⇒ w = 42 ft tan15° = x

x + y = 464 m

92.

95.

5.0 15.0 2.35 mi

y

o

o

24.0

6.0

x

y x + 2.35 y y ° tan 24.0 = ⇒ x = x tan 24.0° y tan15.0° = y ⇒ y = 1.58 mi + 2.35 tan 24.0° tan15.0° =

93.

5.0 θ

d θ

22.5

5.0

2θ + 90° + 22.5° = 180°

θ=

67.5° 2

o

h 7.0

1.0

1.0 = 80.4° 6.0 1.0 ° = 99.6° upper angle = 90 + sin −1 6.0 h 1 A = ( b1 + b2 ) = (1.0 tan 80.4° ) ( 7.0 + 5.0 ) = 35 m 2 2 2 123 base angle = cos −1