Solutions manual for trigonometry 8th edition by mckeague ibsn 9781305652224 by Nelson9475

Chapter 2 Right Triangle Trigonometry

2.1 Definition II: Right Triangle Trigonometry EVEN SOLUTIONS 2. 4. 6.

Using Definition II and Figure 8, we would refer to a as the side opposite A, b as the side adjacent to A, and c as the hypotenuse. a. cosine (ii) b. cosecant (iii) c. cotangent (i) Using the Pythagorean Theorem, first find a: a 2 8 2 17 2 a 2

64 289 a 2 225 a 15 Using a = 15, b = 8, and c = 17, write the six trigonometric functions of A: a 15 b 8 sin A cos A c 17 c 17 c 17 c 17 csc A sec A a 15 b 8 Using the Pythagorean Theorem, first find c: 5 2 22 c2

a 15 b 8 b 8 cot A a 15 tan A

25 4 c 2 c 2 29 c 29 Using a = 5, b = 2, and c 29 , write the six trigonometric functions of A: a 5 5 29 b 2 2 29 sin A cos A c 29 c 29 29 29

tan A

a 5 b 2

csc A

10.

sec A

a 5 Using the Pythagorean Theorem, first find c:

ď&#x20AC;

5 2 11 c2 25 11 c2 c2 36 c 6

29 b

cot A

b 2 a 5

Using a = 5, b 11 , and c = 6, write the six trigonometric functions of A: a 5 b 11 sin A cos A c 6 c 6 c 6 6 11 c 6 sec A 11 a 5 b 11 Using the Pythagorean Theorem, first find a: a 2 32 4 2 csc A

12.

tan A cot A

5 b

5 11 11 11

11 a

a 2 9 16 a2 7 a Using a

7 , b = 3, and c = 4, find the three trigonometric functions of A:

7 b 3 a 7 cos A tan A c 4 c 4 b 3 Now use the Cofunction Theorem to find the three trigonometric functions of B: 3 7 b 3 sin B cos A cos B sin A tan B cot A 4 4 a Using the Pythagorean Theorem, first find c: 32 12 c 2 sin A

14.

3 7 7

9 1 c2 c 2 10

16.

c 10 Using a = 3, b = 1, and c 10 , find the three trigonometric functions of A: a 3 a 3 3 10 b 1 10 tan A 3 sin A cos A c 10 c 10 b 1 10 10 Now use the Cofunction Theorem to find the three trigonometric functions of B: 1 10 3 3 10 b 1 sin B cos A cos B sin A tan B cot A 10 a 10 10 10 Using the Pythagorean Theorem, first find c:

ď&#x20AC; 2

12 5

1 5 c2 c2 6 c Using a = 1, b sin A

6 5 , and c

6 , find the three trigonometric functions of A: cos A

tan A

c 6 c 6 b 5 6 6 5 Now use the Cofunction Theorem to find the three trigonometric functions of B: 5 30 1 6 b sin B cos A cos B sin A tan B cot A 5 6 6 a 6 6

18.

Using the Pythagorean Theorem, first find c: x 2 x 2 c2 c2 2 x 2 c

Using a = x, b = x, and c

x 1 2 b x 1 2 a x cos A tan A 1 c 2 c 2 b x 2x 2 2x 2 Now use the Cofunction Theorem to find the three trigonometric functions of B: 1 2 1 2 b x sin B cos A cos B sin A tan B cot A 1 2 2 a x 2 2 The coordinates of point B are B 8, 6 . Using the Pythagorean Theorem, first find c: sin A

20.

2 x , find the three trigonometric functions of A:

62 82 c2

22. 24. 26. 28. 30. 32.

36 64 c 2 c2 100 c 10 Using a = 6, b = 8, and c = 10, find the three trigonometric functions of A: a 6 3 b 8 4 a 6 3 sin A cos A tan A c 10 5 c 10 5 b 8 4 c c 1 Since b c , 1. Since sec 1 , it is impossible for sec . b b 2 c c Since b c , 1 and can be as large as possible. Since sec , sec can be as large as possible. b b Using the Cofunction Theorem, cos 70 sin 20 . Using the Cofunction Theorem, cot 22 tan 68 . Using the Cofunction Theorem, csc y sec 90 y . Using the Cofunction Theorem, sin 90 y cos y . x 0 30

34.

Complete the table, using the ratio identity sec x

1 : cos x

45 60

36.

Simplifying the expression: 5 sin 60 5

38.

Simplifying the expression: cos 3 60

40.

Simplifying the expression: sin 60 cos 602

2 3 15 3   5 2 4 4

1  3 2

⋅

90

cos x

sec x

3 2

2 2 3 3 3 2 2 2

2 2 1 2 0

2 undefined

1 8  2

3 1 2

2

31    2

42 3 2 3 4 2

1 

42.

Simplifying the expression: sin 45 cos 45

44.

Simplifying the expression: tan 2 45 tan 2 60 12

46.

Simplifying the expression: 6 cos x 6 cos 30 6



2 2

3 13 4 3 2

3 3

48.

Simplifying the expression: 2 sin 90 y 2 sin 90 45 2 sin 45 2

50.

Simplifying the expression: 5 sin 2 y 5 sin 2 45 5 sin 90 5 1 5

52.

Simplifying the expression: 2 cos 90 z 2 cos 90 60 2 cos 30 2

54. 56.

58. 60.

62. 64. 66.

Finding the exact value: csc 30

1 1 2 sin 30 1

3.68 2 b 2 5.932 b 2 5.932 3.68 2

sin B cos A 0.78 68.

2 1 1 Finding the exact value: sec 60 2 cos 60 1 2 3 cos 30 2 Finding the exact value: cot 30 3 1 sin 30 2 1 1 Finding the exact value: csc 45 2 1 sin 45 2 1 1 Finding the exact value: sec 90 , which is undefined cos 90 0 cos 0 1 Finding the exact value: cot 0 , which is undefined sin 0 0 First find a using the Pythagorean Theorem:

b 2 21.6225 b 4.65 Now find sin A and cos A : a 3.68 sin A 0.62 c 5.93 Using the Cofunction Theorem:

cos A

b 4.65 0.78 c 5.93

cos B sin A 0.62

First find c using the Pythagorean Theorem: 13.64 2 4.77 2 c2 c 2 208.8025 c 14.45 Now find sin A and cos A : a 13.64 sin A 0.94 c 14.45 Using the Cofunction Theorem:

sin B cos A 0.33

cos A

0 2 0

b 4.77 0.33 c 14.45

cos B sin A 0.94

70.

Since CG CD 3 , using the Pythagorean Theorem: (CG )2 (CD)2 (DG )2 32 32 (DG )2 9 9 (DG )2 (DG )2 18 DG 18 3 2 Now use the Pythagorean Theorem with DGE : (DG )2 (GE )2 (DE )2 3 2

32 (DE )2

18 9 (DE )2 (DE )2 27 DE 27 3 3 Now, let represent the angle formed by diagonals DE and DG. Therefore: GE 3 1 3 DG 3 2 2 cos DE 3 3 3 DE 3 3 3 Let CG CD x , using the Pythagorean Theorem: (CG )2 (CD)2 (DG )2 x 2 sin

72.

6 3

x 2 (DG )2 (DG )2 2 x2 DG 2 x 2 2 x Now use the Pythagorean Theorem with DGE : (DG )2 (GE )2 (DE )2 2x

x 2 (DE )2

2 x 2 x 2 (DE )2 (DE )2 3x 2 DE 3x 2 3 x Now, let represent the angle formed by diagonals DE and DG. Therefore: GE x 1 3 DE 3x 3 Using the distance formula: sin

74.

ď&#x20AC;

2 x 12 2 5 2 13 2

x 1 9 13 2

x1 4 x 1 2, 2 x 1, 3

cos

DG 2 x 2 DE 3x

6 3

76.

A point on the terminal side is 1,1 . Drawing the angle in standard position:

78. 80.

A coterminal angle to –210° is 150°. Since sin 35° = cos (90° – 35°) = cos 55°, the correct answer is d.  2 1 3 Simplifying the expression: 4 cos2 30 2 sin 30 4 2 4

82.

2 

3 1 3 1 4 . The correct answer is c.

ODD SOLUTIONS 1. 5.

triangle measure 2

c b 2

25 9 16 4 a 4 c 5 b 3 cos A c 5 a 4 tan A b 3



Simplify a 2 2 5 b 1 cot A c 5 5 a 2 c 5 b 1 5 sec A cos A b 1 c 5 5 c 5 a 2 csc A tan A 2 a 2 b 1 Pythagorean Theorem

sin A

11.

Substitute known values

a 2 c 3 b 5 cos A c 3 a 2 2 5 b 5 5

cot A

2 c 3 3 5 sec A b 5 5 csc A

c2 a 2 

6 5

Substitute known values

 36 25  11

Simplify

sin A

Substitute known values

2 1 41 5

b 3 a 4 c 5 sec A b 3 c 5 csc A a 4 Pythagorean Theorem

45 9 3

tan A

cot A

c a2 b2 2



Pythagorean Theorem

Simplify

sin A

complement

c a2 b2

Substitute known values

5 3

3. Pythagorean Theorem

c 3 a 2

Simplify

a 5 c 6 b 11 cos A c 6

sin A

tan A

5 b

5 11 11 11

sin B

11 c

a 5 c 6 b 11 tan B a 5

cos B

c a2 b2

13.

Pythagorean Theorem

Substitute known values

1 1 11

Simplify

a 1 2 c 2 b 1 2 cos A c 2 a 1 tan A 1 b 1 sin A

b 1 2 c 2 2 a 1 2 cos B c 2 2 b 1 tan B 1 a 1 sin B

2 2

2 15.

c2 a 2

Pythagorean Theorem

10 2 6 2

Substitute known values

100 36

Simplify

a 6 3 c 10 5 b 8 4 cos A c 10 5 a 6 3 tan A b 8 4

sin A

b 8 4 c 10 5 a 6 3 cos B c 10 5 b 8 4 tan B a 6 3

sin B

64 8 17.

a c2 b 2 2

2x x

Pythagorean Theorem 2

Substitute known values

4 x2 x2 3x

Simplify

x 3

21.

23.

25.

a 3, b 4, c 5 sin A

a 3

c b 4 cos A c a 3 tan A b

31. csc x

1 sin x undefined

35.

The coordinates of B are (4, 3).

5 5

4 a x 3 3 b x 1 sin A sin B c 2x 2 c 2x 2 a x 3 3 b x 1 cos B cos A c 2x 2 c 2x 2 a x 3 b x 1 3 tan A 3 tan B b x a x 3 3 3 adj side 3 cos For this to be true, the adjacent side would have to be three times larger than the hypotenuse. hyp 1 This is impossible since the hypotenuse is the longest side of a right triangle. opp side The opposite and adjacent sides can be any number greater than 0. If we choose a very large tan adj side number for the opposite side and a very small number for the adjacent side, the ratio will approach infinity. 27.

29. 33.

19.

37. 39. 3 1 4 4 4 4

1 41. 2

2 4

43. 1 31 3



12 3 3 4 2 3 45.

47.

1 3 2 3 2

4 49.

51. 3  2 

3 3 2 2 53.

2 2

Reciprocal identity 1

Substitute exact value from Table 1

3 /2 2 2 3 3

55.

Division of fractions 57.

Ratio identity

2 /2 3 /2

2 2 3 3

Substitute values from Table 1

2 /2 3

Simplify

59.

61.

Ratio identity

1 1/ 2

1 /2 3 /2 1

Substitute values and simplify

1 1

a 3.70 0.38 c 9.62 b 8.88 cos A 0.92 c 9.62 b 8.88 sin B 0.92 c 9.62 a 3.70 cos B 0.38 c 9.62

c2 b 2

sin A

9.62 8.88

13.69 3.70

67.

a 19.44 0.96 c 20.25 b 5.67 cos A 0.28 c 20.25 b 5.67 sin B 0.28 c 20.25 a 19.44 cos B 0.96 c 20.25

c a2 b2

sin A

19.44 5.67

410.0625 20.25 69.

CH 

CD 2 DH

CF 

CH FH 

52 52 25 25 50 5 2 sin

cos

CD 2 DH

x2 x2 2x2 x 2



CH CF 5 2 5 3 2 6 or 3 3

3 3 CH 

5 2

50 25 75 5 3

CF 5 5 3

71.

Reciprocal identity 1

65.

3 3

63.

Substitute values from Table 1

CF 

CH FH 

x 2



2x2 x2 3x 2 x 3

Simplify

sin

73.

cos

CF x x 3

CF x 2 x 3

3 3

x2 x1  y2 y1 

r

3 1

2 4

6 3

Distance formula 2

4 2 22

or 3

Substitute known values Simplify

16 4

20 2 5 75.

The terminal side is the line y x . Some points in quadrant II on the line y x are 1,1,

2, 2, and 3, 3.

77. a 16 4 c 20

The answer is c.

79.

sin A

81.

Statement a is false because

2.2 Calculators and Trigonometric Functions of an Acute Angle EVEN SOLUTIONS

6. 8. 10. 12. 14. 16. 18. 20. 22. 24.

If 7.25 in decimal degrees, then the 7 represents the number of degrees, the 2 represents the number of tenths of a degree, and the 5 represents the number of hundredths of a degree. On a calculator, the SIN–1, COS–1, and TAN–1 keys allow us to find an angle given the value of a trigonometric function. Adding the angles: 1141 3216 4357 Adding the angles: 6338 2452 8790 8830 Adding the angles: 7721 2644 1045 Subtracting the angles: 90 6225 8960 6225 2735 Subtracting the angles: 180 13239 17960 13239 4721 Subtracting the angles: 8938 2858 8898 2858 6040 Converting to degrees and minutes: 83.6 83 0.6 83 0.6(60) 8336 Converting to degrees and minutes: 78.5 78 0.5 78 0.5(60) 7830 Converting to degrees and minutes: 43.85 43 0.85 43 0.85(60) 4351 Converting to degrees and minutes: 8.3 8 0.3 8 0.3(60) 818

26.

Converting to decimal degrees:

28.

Converting to decimal degrees:

30.

Converting to decimal degrees:

32.

Converting to decimal degrees:

34. 36.

Calculating the value: cos 79.2 0.1874 Calculating the value: sin 4 0.0698

2. 4.

38. 40. 42. 44.

Calculating the value: tan 41.88 0.8966 1 Calculating the value: cot 29 1.8040 tan 29 1 Calculating the value: sec18.7 1.0557 cos18.7 1 Calculating the value: csc 77.77 1.0232 sin 77.77

46.

Calculating the value:

48.

Calculating the value:

50.

Calculating the value:

52.

Calculating the value:

54.

x 0 15 30 Completing the table: 45 60 75 90

56.

Finding the angle : sin 1 (0.7139 ) 45.6

58.

Finding the angle : cos1 (0.0945) 84.6

60.

Finding the angle : tan 1(6.2703) 80.9 1 Since sec 8.0101 , cos , so cos1 8.0101 1 Since csc 4.2319 , sin , so sin1 4.2319 1 Since cot 7.0234 , tan , so tan1 7.0234

62. 64. 66.

csc x sec x cot x error (undefined) 1 error (undefined) 3.8637 1.0353 3.7321 2 1.1547 1.7321 1.4142 1.4142 1 1.1547 2 0.5774 1.0353 3.8637 0.2679 1 error (undefined) error (undefined)

1 1 1

82.8 . 8.0101 13.7 . 4.2319 8.1 . 7.0234

68.

Finding the angle : sin 1 (0.9459) 71.0672 71 0.0672(60) 714

70.

Finding the angle : tan 1 2.4652 67.9202 67 0.9202(60) 6755 1 Since sec 1.9102 , cos . 1.9102  1  Finding the angle : cos1 58.4323 58 0.4323(60) 5826 1.9102  Calculating the values: sin13 0.2250 and cos 77 0.2250 Calculating the values: sec 6.7 1.0069 and csc 83.3 1.0069 Calculating the values: tan 3515 tan 35.25 0.7067 and cot 5445 cot 54.75 0.7067 Calculating the value: cos 2 58 sin 2 58 1

72.

74. 76. 78. 80. 82.

To calculate B, B sin 1 (4.321) , which results in an error message. Since, for any angle B, sin B 1, it is impossible to find an angle B such that sin B 4.321 .

84.

86.

undefined value, cot 0 is undefined. 3 2.5 2 x a. Completing the table: cot x 19.1 22.9 28.6

1.5 1 0.5 38.2 57.3 114.6

0.6 0.5 0.4 0.3 0.2 x cot x 95.5 114.6 143.2 191.0 286.5 Using 36.597 and h = 5 in the shadow angle formula: tan sin 36.597 tan(5 15) 2.2250

b. 88.

To calculate cot 0 , we would find tan 0 0 then find the reciprocal. This results in an error message. Since

Completing the table:

0 undefined

0.1 0 573.0 undefined

⋅

tan 2.2250 65.8 90.

First find the value of r: r 

3 12

4 2

Finding the three trigonometric functions using x sin 92.

94.

cos

x r

3 2

tan

Let 1, 1 be a point on the terminal side of –135°. First find the value of r: r 12 (1)2

(4)2 32

16 9

3 3

11 2

25 5

Converting to decimal degrees: 1 1 100. Since cot x , tan . Then tan 1 . The correct answer is a. x x

. The correct answer is b.

ODD SOLUTIONS minutes, seconds

3. 7. since

1 3

Finding the remaining trigonometric functions using x = –4, y = 3, and r = 5: y 3 x 4 x 4 sin cos cot r 5 r 5 y 3 r 5 r 5 csc sec y 3 x 4 Since sec 0 , x > 0. Thus for tan 0 , we must have y < 0. Thus the terminal side of lies in quadrant IV.

98.

1. 5.

y x

Finding the three trigonometric functions using x = –1, y = –1, and r 2 : y 1 x 1 y 1 2 2 sin cos tan 1 x 1 r 2 r 2 2 2 3 Since tan and terminates in quadrant II (where x < 0 and y > 0), choose x = –4 and y = 3. Finding r: 4

96.

y 1 r

3 , y = 1, and r = 2:

11.

value, angle

1 is an 0

13.

Change

to 60 '

15.

17.

19.

21.

23.

25.

27.

29.

31.

33.

Scientific Calculator: 27.2 sin Graphing Calculator: sin ( 27.2 ) ENTER

35.

Answer to 4 places: 0.4571 Scientific Calculator: 18 cos Graphing Calculator: cos ( 18 ) ENTER

37.

Answer to 4 places: 0.9511 Scientific Calculator: 87.32 tan Graphing Calculator: tan ( 87.32 ) ENTER Answer to 4 places: 21.3634

39. Scientific Calculator: 31 tan 1 / x 1 Graphing Calculator: tan ( 31 ) x

ENTER

Answer: 1.6643 41. Scientific Calculator: 48.2 cos 1 / x Graphing Calculator: cos ( 48.2 ) x 1 Answer: 1.5003 43. Scientific Calculator: 14.15 sin 1 / x Graphing Calculator: sin ( 14.15 ) Answer: 4.0906

x 1 ENTER

45. Scientific Calculator: 24.5 cos Graphing Calculator: cos ( 24.5 ) ENTER Answer: 0.9100 47. Scientific Calculator: 42.25 tan Graphing Calculator: tan ( 42.25 ) ENTER Answer: 0.9083 49. Scientific Calculator: 56.67 sin Graphing Calculator: sin ( 56.67 ) ENTER Answer: 0.8355 51.

Scientific Calculator: 45.9 cos 1 / x 1 Graphing Calculator: cos ( 45.9 ) x

53.

55.

ENTER

Answer: 1.4370 Use your calculator to find the values of the sine, cosine, and tangent of each angle: x tan x cos x sin x 0 1 0 0.2588 0.9659 0.2679 0.5 0.8660 0.5774 0.7071 0.7071 1 0.8660 0.5 1.7321 0.9659 0.2588 3.7321 1 0 Error (undefined) Scientific Calculator: 0.9770 inv cos Graphing Calculator: 2nd cos ( 0.9770 ) ENTER

57.

Answer: Scientific Calculator: 0.6873 inv tan Graphing Calculator: 2nd tan ( 0.6873 ) ENTER

59.

Answer: Scientific Calculator: 0.9813 inv sin Graphing Calculator: 2nd sin ( 0.9813 ) ENTER Answer:

61.

sec 1.0191 cos 1  cos

63.

Graphing Calculator: Answer:

1.0191 1 1.0191

csc 1.8214 1 sin  sin

1 1.0191  inv cos 2nd cos ( 1 1.0191 ) ENTER

Scientific Calculator:

1.8214 1 1.8214

Scientific Calculator:

1 1.8214  inv sin

Graphing Calculator:

2nd sin ( 1 1.8214 ) ENTER

Answer:

65.

cot 0.6873

Scientific Calculator:

67.

1 Graphing Calculator: 0.6873 tan 1 Answer: tan 0.6873 Scientific Calculator: 0.4112 inv cos Answer in decimal degrees is Convert the decimal part to minutes:

1 0.6873

 inv tan

2nd tan ( 1 0.6873 ) ENTER

0.719 60

Graphing Calculator: 2nd cos ( 0.4112 ) 2nd APPS

DMS ENTER

Answer: 69.

cot 5.5764 1 tan

5.5764

1 5.5764 Scientific Calculator: 1 5.5764  inv tan tan

Answer in decimal degrees is Convert the decimal part to minutes: 0.1666 60 Graphing Calculator: 2nd tan ( 5.5764 ) 2nd APPS

DMS ENTER

Answer: 71.

csc 7.0683 1 sin 

7.0683

1 7.0683 Scientific Calculator: 1 7.0683  inv sin sin

Answer in decimal degrees is Convert the decimal part to minutes: 0.133 60 Graphing Calculator: 2nd sin ( 1 7.0683 ) 73.

Answer: Scientific Calculator:

23 sin

Graphing Calculator:

sin ( 23 ) ENTER

Both answers should be 0.3907.

and

2nd APPS

DMS ENTER

67 cos cos ( 67 ) ENTER

75.

To calculate Scientific Calculator: 34.5 cos 1 / x Graphing Calculaltor: cos ( 34.5 ) x 1

ENTER

To calculate Scientific Calculator: 55.5 sin 1 / x 1 Graphing Calculaltor: sin ( 55.5 ) x

77.

ENTER

Both answers should be 1.2134. Scientific Calculator: 4.5 tan Graphing Calculaltor: tan ( 4.5 ) ENTER To calculate Scientific Calculator: 85.5 tan 1 / x 1 Graphing Calculaltor: tan ( 85.5 ) x

ENTER

Both answers should be 0.0787. 79.

Scientific Calculator: 37 cos x 2

2 37 sin x ď&#x20AC;

Graphing Calculator: cos ( 37 ) x 2

) x 2 ENTER

sin ( 37

81.

Answer should be 1. Scientific Calculator: 1.234 inv sin

83.

Graphing Calculator: 2nd sin 1.234 ENTER You should get an error message. The sine of an angle can never be greater than 1. Scientific Calculator: 90 tan Graphing Calculaltor: tan ( 90 ) ENTER

87.

You should get an error message. The tangent of is undefined. where and h 2

.333478 tan1 .333478 ď&#x20AC; 89.

x, y 3, 2

sin

y 2 2 1313 r

x 3 and y 2 r

32 2 94

cos tan

3 13 r 13

y 2 x

13 2 3

91.

A point on the terminal side of an angle of x = 0, y = 1, and r = 1.

in standard position is (0, 1), where

is undefined 93.

5 and is in QIII. In QIII, both x and y are negative. 13

cos x 5 r

cos

x 5 and r 13 x 2 y2 r 2 2

5 y 13 2

25 y 169

y 12

sin

y 12 12 x 5 x 5 5 cot y 12 r 13 sec x 5 tan

y 2 144

csc

r 13 y 12

12 13 5 12 13 5 13 12

y 12 y 12 because is in QIII 95.

97.

The sin is positive in QI and QII. The cos is negative in QII and QIII. Therefore, must lie in QII. Change

to 60 '

The answer is d.

2.3 Solving Right Triangles EVEN SOLUTIONS 2. a 4. 6.

If the sides of a triangle are accurate to three significant digits, then angles should be measured to the nearest tenth of degree, or the nearest ten minutes. In general, round answers so that the number of significant digits in your answer matches the number of significant digits in the least significant number given in the problem. a. three b. three c. five d. three a. five b. five c. five d. seven

10.

Begin by drawing ABC :

Therefore: sin 52

12.

15 a 15 sin 52 12 ft Begin by drawing ABC :

Therefore: 55 c c cos 64 55 cos 64

55 64ď&#x20AC;

c 14.

125 m 130 m cos

Begin by drawing ABC :

Therefore: a

cos 24.5

16.

3.45 a 3.45 cos 24.5 3.14 ft Begin by drawing ABC :

Therefore: b 43.21 b 43.21 tan17.71 13.80 in.

tan17.71

18.

Begin by drawing ABC :

Therefore: tan A

16 29

 29 29  Begin by drawing ABC : A tan 1

20.

Therefore: cos A

22.

2.7 7.7

 2.7 A cos1  69 7.7  Begin by drawing ABC :

Therefore: 2.345 5.678 2.345  24.39 A sin 1 5.678  Begin by drawing ABC and label missing information: sin A

24.

Note that B 90 71 19 . Therefore: a sin 71 36 a 36 sin 71 34 m

cos 71

36 b 36 cos 71 12 m

26.

Begin by drawing ABC and label missing information:

Note that B 90 48.3 41.7 . Therefore: 3.48 sin 48.3 c c sin 48.3 3.48 3.48

28.

4.66 in. sin 48.3 Begin by drawing ABC and label missing information:

Note that B 90 6654 8960 6654 236 . So: 88.22 cos 6654 c c cos 6654 88.22

cos 48.3

4.66 b 4.66 cos 48.3 3.10 in.

a 88.22 a 88.22 tan 6654 206.8 cm

tan 6654 

30.

88.22 224.9 cm cos 6654 Begin by drawing ABC and label missing information:

32.

Note that A 90 21 69 . Therefore: b a cos 69 sin 69 4.2 4.2 b 4.2 cos 69 1.5 ft a 4.2 sin 69 3.9 ft Begin by drawing ABC and label missing information:

Note that A 90 5330 8960 5330 3630 . So: 125 cos 3630 c c cos 3630 125

a 125 a 125 tan 3630 92.5 mm

tan 3630 

125 156 mm cos 3630 Begin by drawing ABC and label missing information: c

34.

Note that A 90 44.44 45.56 . Therefore: 5.555 sin 45.56 c c sin 45.56 5.555

5.555 b b tan 45.56 5.555

5.555 7.780 mi sin 45.56 Begin by drawing ABC and label missing information: c

36.

Therefore: c tan A

85 2 992 130 ft . Finding the angles: 99 85

99   49 85  B 90 49 41 Begin by drawing ABC and label missing information: A tan 1

38.

Therefore: b

73.6 2 62.32 39.2 cm . Finding the angles:

62.3 73.6 62.3  57.8 A sin 1  73.6 B 90 57.8 32.2

sin A

tan 45.56

5.555 tan 45.56

5.447 mi

40.

42.

Since the right triangle is a 45°–45°–90° triangle, its height is 2.0. Therefore: 2.0 tan A 3.0 2.0 A tan 1 34 3.0 Re-drawing the figure: 44. Re-drawing the figure:

Using the right triangle:

Using the right triangle: cos 45

cos 62

19 19 x

19 x cos 62 19 19 x

46.

19 cos 62 19 x 19 21 cos 62

Re-drawing the figure:

First find x:

48.

r 15 (r 15) cos 45 r r cos 45 15 cos 45 r 15 cos 45 r r cos 45 15 cos 45 r 1 cos 45 15 cos 45 r 1 cos 45 2 15   2  15 2 r 2 2 2 1 2 Re-drawing the figure:

First find x: x tan 48 24 x 24 tan 48 27

sin 49

19 x 19 sin 49 14

Now find h:

Now find ABD :

tan 53

32 14

tan ABD 

47 h 27 tan 53 35

ABD tan 1

32 14

50.

Re-drawing the figure:

52.

First find x:

Re-drawing the figure:

First find h:

cos 48

sin 32

12 x 12 cos 48 8.0

Now find x: 8y

30 x x tan 48 30 tan 48

17 8 y 17 cos 23 y 17 cos 23 8 7.6

54.

Re-drawing the figure:

, so h x tan 61 . Also note that tan 32 x two expressions equal: x tan 61 (14 x) tan 32 x tan 61 14 tan 32 x tan 32 x tan 61 x tan 32 14 tan 32 x tan 61 tan 32 14 tan 32 14 tan 32 tan x 7.4 61 tan 32 Note that tan 61

56.

56 h 56 sin 32 30

Now find y: cos 23

Since GC CD 3.00 , using the Pythagorean Theorem: GD GE 3 1 tan GDE GD 3 2 2 1 GDE tan 1 35.3 2

30 27 tan 48

, so h (14 x) tan 32 . Setting these 14 x

32 32

18 3 2 . Therefore:

58.

First find CAB : tan CAB

Now find EAB : 66

CAB tan 60.

62. 64.

66.

68.

70.

54 1 66

tan EAB 50.71 54

EAB tan

54 1 78

55.30 54

Therefore: CAE EAB CAB 55.30 50.71 4.6 Let O represent the center of the goal. First find OAD : Now find OAF : 6 12 tan OAD tan OAF 54 54 6 12 OAD tan 1 6.34 OAF tan 1 12.53 54 54 Therefore: DAF OAF OAD 12.53 6.34 6.2 Since CAE is also 6.2°, the sum of the angles is 12.4°. Since 12.4° is much greater than 4.6°, the chance of scoring is much better from the center than from the corner of the penalty area. From Example 5, we have: 139 h cos 135 125 1 139 h 125 2 125 139 h 2 125 h 139 227.4 2 The rider is approximately 230 ft above the ground. First note that the distance from the ground to the low point of Colossus is 174 – 165 = 9 ft. The radius is 82.5 ft. Since x h 82.5 9 91.5 , x 91.5 h . Therefore: x 91.5 h cos 82.5 82.5 91.5 h 82.5 cos  h 91.5 82.5 cos  a. Substituting 150 : h 91.5 82.5 cos150 163 ft b. Substituting 240 : h 91.5 82.5 cos 240 133 ft c. Substituting 315 : h 91.5 82.5 cos 315 33.2 ft First note that the distance from the ground to the low point of the High Roller is 550 – 520 = 30 ft. The radius is 260 ft. Since x h 260 30 290 , x 290 h . Therefore: x 290 h cos 260 260 290 h 260 cos h 290 260 cos Substituting 110 : h 290 260 cos110 379 ft 380 ft 7 Y1 2 Entering the functions Y1 X 80 70 and Y2 tan 1 , complete the table: 640 X X 10 5 1 0.5 0.1 0.01 Y1 16.4063 8.4766 1.7391 0.8723 0.1749 0.0175 Y2

58.6

59.5

60.1

60.2

60.3

Based on these results, it appears the angle between the cannon and the horizontal is approximately 60.3°.

72. 74.

Since csc B 5 , sin B

, so sin 2 B 5

1

 5

3

2 Finding cos A : cos 2 A 1 sin 2 A 1

1 . 25 2

4

9 7 16 16

7 . Since A terminates in quadrant II, where x < 0, cos A 0 . Thus cos A 4 First find sin (note sin 0 since terminates in quadrant IV): 2 1 4 2 5 2 sin 1 cos 1 5 5 5 5 Now find the other four trigonometric ratios using x = 1, y = –2, and r 5 : So cos A

76.

7 4

1 sec 78.

r x

82.

5 2

tan

2 x

1

2

1 1

3 4

3 2

Now find the other three trigonometric ratios using x 3 , y = –1, and r = 2: 22 3 r y 1 1 3 x 3 cot  sec tan  3 y 1 x 3 x 3 3 3 3 Finding side b: b cos A c b cos 58 15 b 15 cos 58 7.9 ft The correct answer is c. Let x represent the height of the rider above the center of the wheel (which is 51.5 feet above the ground). Since the point of the rider is 50° above the horizontal, we have: h 51.5 sin 50 45 h 51.5 45 sin 50 h 51.5 45 sin 50 86 ft The correct answer is c.

ODD SOLUTIONS 1. 5.

cot

1 Since csc 2 , sin . Now find cos (note that cos 0 since terminates in quadrant III): 2 cos 1 sin 2

80.

csc

left, right, first nonzero, not a. 2 b. 3 c. 2 d. 2

3. 7.

sides, angles a. 4 b. 6 c. 4

d. 4

Cosine relationship Multiply both sides by 89

89 0.7431 66 cm

Substitute value for Answer rounded to 2 significant digits

11.

Sine relationship Multiply both sides by c Divide both sides by 22 0.5592 c 39 m c

13.

Substitute value for Answer rounded to 2 significant digits Sine relationship

7.55 0.2907 2.19 cm 15.

Multiply both sides by 7.55 Substitute value for Answer rounded to 3 significant digits Tangent relationship Multiply both sides by a Divide both sides by

12.34 1.4458 a 8.535 yd 32.4 tan B 42.3 a

17.

0.7659 B tan1 0.7659

Substitute value for Answer rounded to 4 significant digits Tangent relationship Divide Use calculator to find angle Answer rounded to the nearest tenth of a degree

19.

9.8 sin B 12

0.8166 B cos 1 0.8166

Sine relationship Divide Use calculator to find angle Answer rounded to the nearest degree

21.

23.32 cos B 45.54

0.5120 B cos 1 0.5120

Cosine relationship Divide 23.32 by 45.54 Use calculator to find angle Answer rounded to the nearest hundredth of a degree

23.

First, we find B : Next, we find side a: Sine relationship

a 10 m

Multiply both sides by 24 Answer rounded to 2 significant digits

Last, we find side b: Cosine relationship

b 22 m

Multiply both sides by 24 Answer rounded to 2 significant digits

25.

First, we find B : Next, we find side c: Sine relationship Multiply both sides by c then divide by

80.6 in Last, we find side b:

Answer rounded to 3 significant digits Tangent relationship

27.

Multiply both sides by 43.4 Answer rounded to 2 significant digits

67.9 in First, we find B : Next, we find side a:

Tangent relationship Change angle to decimal degrees Multiply both sides by 5.932 Answer rounded to 4 significant digits

a 1.121 cm Last, we find side c:

Cosine relationship

29.

Multiply both sides by c then divide by Answer rounded to 4 significant digits

c 6.037 cm First, we find A : Next, we find side a:

Cosine relationship Multiply both sides by 5.8 Answer rounded to 2 significant digits

1.4 ft Last, we find side b:

Sine relationship

5.6 ft

Multiply both sides by 5.8 and round to 2 significant digits

31.

First, we find A : Next, we find side a: Tangent relationship Change angle to decimal degrees

a 650 mm

Multiply both sides by a then divide by Answer rounded to 3 significant digits

Last, we find side c: Sine relationship Multiply both sides by c then divide by

726 mm 33.

Answer rounded to 3 significant digits

First, we find A : Next, we find side b: Tangent relationship

2.356 mi

Multiply both sides by 5.432 Answer rounded to 4 significant digits

Last, we find side c: Cosine relationship Multiply both sides by c and then divide by

5.921 mi 35.

First, we find A : 37 tan A 87

0.4253 A tan 1 0.4253

Answer rounded to 4 significant digits Tangent relationship Divide 37 by 87 Use calculator to find angle Answer rounded to nearest degree

Next, we find B : Last, we find c:

c 2 37 2 87 2 1369 7569 8938 c 95 95 ft

Pythagorean Theorem Simplify Simplify Take square root of both sides c must be positive

37.

First, we find A : 377.3 cos A 588.5

0.6411 A cos1 0.6411

Cosine relationship Divide Use calculator to find angle Answer rounded to nearest hundredth of a degree

Next, we find B :

Last, we find side a:

39.

a 2 377.32 588.5 2 a 2 203, 976.96 a 451.6 451.6 in Using BCD , we find BD:

Pythagorean Theorem Subtract and simplify Take square root of both sides a must be positive Sine relationship

3 Next, we find A 3 sin A 4.0

0.75 A sin1 0.75

Multiply both sides by 6 Exact answer Sine relationship Divide Use calculator to find angle Answer rounded to the nearest degree

41.

Sine relationship Multiply both sides by x + 12 Divide both sides by Subtract 12 from both sides and round to 2 significant digits

43.

Sine relationship Multiply both side by r + 22 Use distributive property Subtract r

from both sides

Factor left side Divide both sides by

Answer rounded to 2 significant digits

45.

Using ABC , we find side x: Tangent relationship

Multiply both sides by 42 Answer rounded to 2 significant digits

Next, using ABD , we find side h: Tangent relationship h 79

h 40 47.

Substitute value for x Multiply both sides by 79 Answer rounded to 2 significant digits

Using ABC , we find side x: Sine relationship

Multiply both sides by 32 Answer rounded to 2 significant digits

Next, using ABD , we find ABD : h Tangent relationship tan ABD x 19 Substitute known values 21 Divide 19 by 21 0.9047 ABD tan1 0.9047 Use calculator to find angle Answer rounded to the nearest degree 49.

Using BCD , we find side x: Cosine relationship

x 7.4

Multiply both sides by 14 Answer rounded to 2 significant digits

Next, using ABC , we find y: Cosine relationship

x y 13.58 7.4 y 13.58 y 6.18 6.2

Multiply both sides by 18 Evaluate right side Substitute value for x Subtract 7.4 from both sides and round to 2 significant digits

51.

Using ABC , we find side h: Sine relationship Multiply both sides by 28 Answer rounded to 2 significant digits

18 Next, using BCD , we find side x:

Tangent relationship Substitute value found for h

53.

Solve for x and round to 2 significant digits Since h is in both ABC and BCD , we will solve for h in the two triangles: In BCD ,

Tangent relationship Multiply both sides by x

In ABC ,

Tangent relationship Multiply both sides by x + y Substitute value for y

Therefore,

Property of equality Distribution Property Subtract

from both sides

Factor left side

55.

Divide both sides by From Problem 69 in Problem Set 2.1, we found that 1 sin 3

0.5774 sin1 0.5774 ď&#x20AC; 57.

59.

From Problem 69 in Problem Set 2.1, we found that 2 cos 3

0.8165 cos 1 0.8165 ď&#x20AC; We know that EC DF 6 ft , EB = 78 ft, CB = 72 ft, DB = 60 ft, and 78 tan EAB 54

tan CAB

1 78 EAB tan 54

EAC EAB CAB

72 54

1 72 CAB tan 54

and

Therefore, the sum of the angles is

DAF DAB FAB .

tan DAB

. 60

DAB tan

54 1 60 54

63. Solve for h

201.5 200 ft 65.

Round to 2 significant digits

r 98.5 a.

h 12 98.5 x x

60.0o 120.0o

49.25 h 12 98.5 49.25 159.8 160 ft h 12 98.5 x

98.5

12 30.0o x

85.3 h 12 98.5 85.3 195.8 196 ft c.

150.0o 98.5

r 12 98.5 12 110.5 h 110.5 x

69.7 h 110.5 69.7 40.8 ft 67.

The radius of the London Eye is cos

71.

x 45.0o

135 67.5 . 2

67.5 44.5 67.5 23 67.5

sec 2 cos

cos

1 sec  1 2 2 1 1   2 4

Reciprocal identity Substitute known value Square both sides

73.

cos 1 sin 2

Pythagorean identity, in QIII 2 2

1  3 5 9 75.



9 5 3

cos 1 sin 2

Pythagorean identity, in QII 

Simplify 1 2 Ratio identity

Reciprocal identity

sin 

1 2 2 3 3 3/2 3

Simplify

Reciprocal identity

cos  1 2 1/ 2

Substitute known values

Substitute value for cos

3 4

3 2

cot

Substitute values

Reciprocal identity tan



2

Ratio identity

Pythagorean identity, in QIII

1 

3 sin  tan cos  3/2 1/ 2

Reciprocal identity

sin 1 cos 2

Reciprocal identity

tan  1 3

sec  1 2

cot 1

csc

Substitute known values

cos

Substitute known value

3 4

1 4 sin  tan cos  3/2 1/ 2

sec

2  1

77.

Simplify

3 3

csc

1 sin 

Reciprocal identity 2

2 3

1 3/2 81.

tan B

35 58

The answer is b.

2.4 Applications EVEN SOLUTIONS 2. 4. 6.

If an observer positioned at the vertex of an angle views an object in the direction of the non-horizontal side of the angle, then this side is called the line of sight of the observer. The bearing of a line is always measured as an angle from the north or south rotating toward the east or west. Sketching a figure: 8. Sketching a figure:

10.

Sketching a figure:

14.

Call x the length of each side. Note the figure:

Therefore: 12.3 x x sin 60 12.3 sin 60

12.3 14.2 sin 60ď&#x20AC; The length of each side is 14.2 in. x

12.

Sketching a figure:

16.

Call w the length of the shorter side (width), and s, t the required angles. Note the figure:

Find w using the Pythagorean Theorem: 278 2 w 2 348 2

18.

w 2 348 2 278 2 43820 w 209 Now find angles s and t: 278 cos s 348 278 s cos 1 37.0 348 t 90 37.0 53.0 The shorter side is 209 mm, and the two angles are 37.0° and 53.0°. Let h represent the height of the hill. Draw the figure:

Therefore: sin 6.5

20.

2.5 h 2.5 sin 6.5 0.28 The hill is approximately 0.28 mi high, which is approximately 1,480 feet. Let s represent the angle between the ladder and the wall. Draw the figure:

Therefore: tan s

4.5 8.5

4.5   28 8.5  The angle between the ladder and the wall is approximately 28°. s tan 1

22.

Let h represent the height of the building. Draw the figure:

Therefore: h

tan 73.4

24.

37.5 h 37.5 tan 73.4 126 The height of the building is approximately 126 feet. Draw the figure with the associated labels:

The sum x y represents the width of the sand pile. Using the two triangles: 15 15 tan17 tan 29 y x x tan 29 15 y tan17 15 15 15 27.1 y 49.1 tan 29 tan17 The width of the sand pile is therefore 27.1 + 49.1 ≈ 76 feet. 5 5 a. First note that in. 1600 1000 ft , which is the horizontal distance between Stacey and Travis. 8 8 b. There are 8 contour intervals between Stacey and Travis, which corresponds to a vertical distance of 8 40 320 ft . c. Let s represent the elevation angle. Construct the triangle: x

26.

⋅

Therefore: tan s

320 0.32 1000

s tan 1 (0.32) 17.7 The elevation angle from Travis to Stacey is approximately 17.7°.

28.

Construct the figure:

First consider the triangle:

Now consider the triangle:

Therefore:

60.0 tan 63.2 x x tan 63.2 60.0 60.0 30.3 ft tan 63.2 The building next door is approximately 39.2 feet tall. Construct the figure: x

30.

60 h 30.3 60 h 30.3 tan 34.5 h 60 30.3 tan 34.5 39.2 ft

tan 34.5

To find his distance from the starting point, use the Pythagorean Theorem: d 2 3.5 2 2.32 17.54 d 17.54 4.2 mi Now find angle s: 3.5 tan s 2.3 3.5 s tan 1 57 2.3 His bearing is S 88° W.

32.

Construct the figure:

Therefore: tan 18.2

34.

21.0 d 21.0 tan 18.2 6.90 The distance from the tree to the rock is 6.90 yards. Construct the figure:

Therefore: sin 6350

36.

111 e 111sin 6350 99.6 mi The boat travels 48.9 mi south and 99.6 mi east. Draw the figure:

cos 6350

s 111 s 111cos 6350 48.9 mi

From the smaller triangle: h tan 65 x x tan 65 h x

h tan 65

From the larger triangle: h tan 54 25 x (25 x ) tan 54 h 25 x x

Setting these two expressions equal: h h 25 tan 65 tan 54 h cot 65 h cot 54 25 h cot 65 h cot 54 25 h cot 65 cot 54 25

h tan 54 h 25 tan 54

25 96 cot 54 cot 65 The height of the obelisk is approximately 96 feet. First find the length CB: 426 tan12.3 CB CB tan12.3 426 h

38.

426 1, 954 ft tan12.3

Therefore: sin BCA

40.

CB AB sin 57.5 1954 AB 1954 sin 57.5 1, 650 ft A rescue boat at A will have to travel approximately 1,650 feet to reach any survivors at point B. Construct a figure:

First note that each person is

35 ft 24.7 ft from the base of the tree. Let h represent the height of the tree. Now 2

construct the triangle:

Therefore: tan 58

42.

24.7 h 24.7 tan 58 40 ft The tree is approximately 27 feet tall. Construct the figure (not drawn to scale):

Using the Pythagorean Theorem: x 2 3960 2 3964.55 2 x 2 15, 681, 600 15, 717, 656.7 x 2 36, 056.7 x 190 The plane is 190 miles from the horizon. Now find angle A: 3960 sin A 3964.55 3960 ď&#x20AC; 87.3ď&#x20AC; A sin1 3964.55

44.

Construct the figure:

Consider the two triangles:

Therefore: tan 75

y 2.5 x y (2.5 x) tan 65

tan 65

x y x tan 75 Setting these two expressions equal: (2.5 x) tan 65 x tan 75 2.5 tan 65 x tan 65 x tan 75 2.5 tan 65 x tan 75 x tan 65 2.5 tan 65 x tan 75 tan 65 x

2.5 tan 65 75 tan 65

0.91

tan

Therefore: 2.5 x 2.5 0.91 1.59 d d d cos 65 1.59 cos 65

1.59 3.8 mi cos 65 Tim is approximately 3.8 miles from the missile when it is launched. 1 1 1 1 Note that sin 1  , sin 2  , sin 3  , ... , thus sin n  . 2 3 4 n1 a. Let x represent the height that is illuminated on the floor. Then: 4 tan 84 x 4 x 0.42 tan 84 The illuminated area is then: 0.42 6.5 2.7 ft 2 . d

46. 48.

Following the procedure from part a: 4 tan 37 x 4 x 5.31 tan 37 The illuminated area is then: 5.31 6.5 34.5 ft 2 . The area is much larger on the winter day.

⋅ sin

54.

2 2 1 sin cos sin  sin  sin 1 sin sincos Working from the left side: cos csc tan cos 1 sin  cos sin cos Working from the left side: 1 cos 1 cos 1 cos cos cos 2 1 cos 2 sin 2

56.

Working from the left side: 1

50. 52.

58.

60.

Simplifying:

1 sin sin 

sin sin 

cos cos 1 1 sec 

⋅

1 cos2 sin 2

cos Let h represent the height of the flagpole. Then: h tan 74.3 22.5 h 22.5 tan 74.3 80.0 The flagpole is 80.0 feet tall. The correct answer is d. Construct a figure:

First note that each person is

35 ft 24.7 ft from the base of the tree. Let h represent the height of the tree. Now 2

construct the triangle:

Therefore: tan 65

24.7 h 24.7 tan 65 53 ft The tree is approximately 53 feet tall. The correct answer is a.

ODD SOLUTIONS 1. elevation, depression 3. north-south For problems 5 through 11, see diagrams in textbook answer section. 13. To find the height, h, we can use the Pythagorean Theorem: 2

h 2 16 42 h 2 256 1, 764 h 2 1, 508 42

h 1, 508 39 cm

To find angle , we can use the cosine ratio: 16 cos 42

 16

The height is 39 cm and the two equal angles are 68 . 15.

Consider the right triangle with sides of 25.3 cm and 5.2 cm (one-half of the diameter): 25.3 tan 5.2 The angle the side makes with the base is . 4.8654

tan1 4.8654 

17.

To find the length of the escalator, x, we use the sine ratio: x 21 ft

21 39 ft 0.5446 The length of the escalator is 39 feet. 19.

sin



43.2 72.5

72.5

0.5959 sin1 0.5959 

21.

43.2

The angle the rope makes with the pole is We use the tangent ratio to find the angle of elevation to the sun, : 73.0 tan 51.0

1.4313 tan 1 1.4313 The angle of elevation to the sun is

73.0 ft

 .

51.0 ft

23. x



29 cm

11o 12o

150 y

25.

32 cm The vertical dimension of the mirror is x + y or 61 cm. a. horizontal distance = 0.50(1,600) = 800 ft b. vertical distance (number of contour intervals)(40) 5 40  200 ft

tan

vertical distance horizontal distance 200 800

0.25 tan1 0.25  27. 9.8 5.9 1.6643

9.8 x

59o 47o

5.9 1.0724 6.3 ft 29.

The vertical dimension of the door is 6.3 feet. We use the Pythagorean Theorem to find the distance x:

x 2 25 2 18 2 625 324 949 x 31 mi

We use the tangent relationship to find angle : 18 tan 25

42o



42o 48o 18

0.72 tan1 0.72  To find the bearing we add bearing is N E.

. The boat is 31 miles from the harbor entrance and its

31. Nipomo

18 2.1445 ď&#x20AC; 39 mi

65o 18

The distance from Lompoc to Buellton is 39 miles.

Lompoc

33.

Buellton

We will call the west distance, x and the north distance, y: x

48.0 mi

79.5

63.4 mi

37 10 '

The boat has traveled 48.0 miles west and 63.4 miles north. 35.

Therefore,

42.17o 33

47.5o x

161 ft The person at point A is 161 feet from the base of the antenna. 37.

24.8 16.8319 417.431

86.6o B 24.8

417.431 0.18895 ď&#x20AC; 78.9 ft The tree is 78.9 feet high.

10.7

39.

First, we will find each person’s distance from the pole, x, using the Pythagorean Theorem:

x 2 x 2 25 2 2 x 2 625 x 2 312.5 x 17.678 ft Next, we will find the height of the pole, h, using the tangent relationship:

26 ft The height of the pole is 26 feet. 41.

112 0.9728  1 0.9728 108.9509 0.02722 4, 000 mi The radius of the earth is 4,000 miles. We want to find x and y in terms of h 

43.

o x 53

h 15 y 31o

We know that x + y = 15. Therefore,

h 0.7536 1.6643 15 2.4179 h 15 h

15 6.2 mi 2.4179

The ship is 6.2 miles from the shore.

45.

1 1

tan1 

tan 2 

1 1

tan

1 2

1 3

tan 3 

0.7071 1 tan 0.7071

0.5774 tan 0.5774

sin cos sin cos sin cos  sin 2 2 sin cos cos 2  2

49.

sin 2 cos 2 2 sin cos 51.

1 2 sin cos cos sin cot sin sin  sincos  sin 

cos

Ratio identity Multiplication of fractions Division of common factor

53.

sec cos  sin  tan  cos  1 cos cos  sin  1 sin 

Reciprocal and ratio identity

Division of fractions Multiplication of fractions and divide common factor

csc 

55.

sec cos

Reciprocal identity cos cos 1 cos L.C.D. is cos cos cos  cos 1 cos 2 Subtraction of fractions cos  2 sin  Pythagorean identity cos 

2.5 Vectors: A Geometric Approach EVEN SOLUTIONS 2. 4. 6. 8. 10.

Two vectors are equivalent if they have the same magnitude and direction. A vector is in standard position if the tail of the vector is placed at the origin of a rectangular coordinate system. If V makes and angle with the positive x-axis when in standard position, then Vx V cos and Vy V sin . If a constant force F is applied to an object and moves the object in a straight line a distance d at an angle with the force, then the work performed by the force is found by multiplying F cos and d. Sketching the vector: 12. Sketching the vector:

14.

Sketching the vector:

16.

Sketching the vector:

18.

Construct the figure for their position after 2 hours (multiply their rates by 2):

Using the Pythagorean Theorem: d 550 2 510 2 750 miles To find the bearing from P1 to P2 , first find the vertical (north-south) change in their positions. This is given by:

550 sin 45 510 sin 45 28.28 miles Construct the triangle:

Therefore: 28.28 750 28.28 87.8 s cos1  750 The bearing from P1 to P2 is S 87.8° W. cos s

20.

Computing the magnitudes of Vx and Vy : Vx 17.6 cos 72.6 5.26 Vy 17.6 sin 72.6 16.8

22.

Computing the magnitudes of Vx and Vy : Vx 383cos1220 374 Vy 383sin1220 81.8

24.

Computing the magnitudes of Vx and Vy : Vx 84 cos 90 0 Vy 84 sin 90 84

26.

Using the Pythagorean Theorem: V

54.2 2 14.5 2 56.1

28.

Using the Pythagorean Theorem: V

2.2 2 8.8 2 9.1

30.

Construct the triangle:

Therefore: sin1.9

32.

135 x 135 sin1.9 4.48 miles The plane will be approximately 4.48 miles off course. Computing the magnitudes of Vx and Vy : Vx 1, 800 cos 55 1, 032 ft 1,sec 000 ft Vy 1, 800 sin 55 1, 474 ft 1,sec 500 ft

34. 36.

sec sec

The horizontal distance traveled is 1.5 1, 032 1, 548 feet 1, 500 feet . Draw the figure corresponding to t = 3 hours:

The west and south distances are given by: west: 960 cos 55 550 km 38.

Using the Pythagorean Theorem: V The elevation angle is given by: 24.3 tan 16.5 24.3 ď&#x20AC; 55.8ď&#x20AC; tan1 16.5

south: 960 sin 55 790 km 2

16.5 24.3

29.4 ft/sec

40.

Construct the figure:

42.

The total distance south and east is given by: east: 68 cos 78 110 cos 30 110 miles The corresponding force diagram would be:

44.

The corresponding force diagram would be:

south: 68 sin 78 110 sin 30 120 miles

Therefore: sin12.5

46.

25.0 F 25.0 sin12.5 5.41 lb The corresponding force diagram would be:

cos12.5

25.0 N 25.0 cos12.5 24.4 lb

Therefore: sin 50

2, 200 T

T sin 50 2, 200

2, 200 H

tan 50

H tan 50 2, 200

50.

2, 200 2, 200 2, 900 lb H 1, 800 lb sin 50 tan 50 The horizontal portion of the force is given by: Fx F cos 35 15 cos 35 lb The work is then given by: Work 15 cos 35 52 640 ft-lb The horizontal portion of the force is given by: Fx F cos15 85 cos15 lb

52.

The work is then given by: Work 85 cos15 110 9, 000 ft-lb Drawing the angle in standard position:

48.

Since r = 1, sin(270) 1, cos(270) 0 , and tan(270) is undefined.

(1)2 12

54.

Choose 1,1 as a point on the terminal side of . Then r

56.

x 1 y 1 2 cos r 2 r 2 2 x 3 6 Since cos , choose x = –6 and r = 10. Now find y: r 5 10 2 2 2 (6) y 10 sin

2 . Therefore: 2 2

36 y 2 100 y 2 64 y8 58.

60.

Using the Pythagorean Theorem: V 9.6 2 2.32 9.9 Finding the angle: 2.3 tan 9.6 2.3  tan1 13 9.6  The correct answer is c. The horizontal portion of the force is given by: Fx F cos 35 28 cos 35 lb The work is then given by: Work 28 cos 35 150 3, 400 ft-lb The correct answer is b.

ODD SOLUTIONS scalar, vector 1. 3. resultant, diagonal 5. horizontal, component, vertical, component 7. zero, static equilibrium For problems 9 through 15, see textbook answer section for diagrams. 17. The first hour, the distance traveled is (9.50 mph)(1 hr) = 9.50 miles The next hour and a half, the distance traveled is o (8.00 mph)(1.5 hr) = 12.0 miles 9.50 mi 37.5 52.5o We will use the Pythagorean Theorem to find x: 12.0 mi 2 2 2 o

x 9.50 12.0 x 2 234.25 x 15.3 mi

37.5

 x

We will use the tangent ratio to find and then add 12.0 tan 9.50

1.2632 tan1 1.2632  The balloon is 15.3 miles from its starting point. The bearing is N 19.

Vx V cos

Vy V sin

21.

12.6 Vx V cos

5.66 Vy V sin

23.

425 0.8073 343 Vx V cos

425 0.5901 251 Vy V sin

64 1 64 25.

35.0 26.0

64 0 0 2

1, 225 676 1, 901 43.6 29.

27.



4.5 3.8

20.25 14.44 34.69 5.9

To find the distance, x, the plane has flown off its course, we can use the sine ratio: 28 mi o

2.8

1.37 miles

true course

31.

Vy V sin

Vx V cos

1200 0.7071 850 feet per second

1200 0.7071 850 feet per second 33.

In 3 seconds, the bullet travels 3(850 ft/sec) = 2,550 ft.

35.

87 37.

The ship has traveled 97 km south and 87 km east. We are given that Vx 35.0 and Vy 15.0 V

35.0 15.0

48o

tan

15.0 35.0 0.4285



1, 225 225 1, 450 38.1 feet per second 39.

Therefore, the velocity of the arrow is 38.1 feet per second at an elevation of . To find the total distance traveled north, we must find the sum of Vy and Wy and to find the total distance traveled east, we must find the sum of Vx and Wx . We are given that V is 170 mi. at an angle of inclination of also that W is 120 mi. at an angle of inclination of

and

or Wx Wy W

53 mi

49o Vx 41o

162 mi Vy

18o

90 mi 80 mi Vy Wy 162 80 242 and Vx Wx 53 90 143

72o

The total distance north is 240 miles and the total distance east is 140 miles, rounded to 2 significant digits. 41.

W 42.0

W 45.0 T

59.4 lb.

42.0 lb.

43.

We are given that W 8.0 W

7.7 pounds 45.

2.1 pounds

W 42.0

33.1 lb 47.

lb, and d 75 ft

Fx F cos

Work Fx

Fx F cos

lb, and d 350 ft Work Fx

d 2900 ft - lb.

49.

x, y 1, 1ď&#x20AC; x 1, y 1 and r

7, 600 ft - lb. 2 1, 1 45o

53.

55.

o W 52.0

51.

15o

A point on the line y = 2x in quadrant I is (1, 2). x = 1, y = 2, and r y 2 2 5 x 1 5 sin cos r 5 r 5 5 5 y 4 8 sin r 5 10 y 8 and r 10

x 2 y2 r 2 2

x 2 8 10 2 x 2 64 100 x 2 36 x6

135o

12 2 2

Chapter 2 Test 1.

First draw the triangle:

Note that c sin A

2 2 12 1

5 . Therefore: 5

cos A

2 2 5 5 5 First draw the triangle: sin B cos A

Note that a sin A

7 2 62

cos A 6

5 2 32

5 5

1 2

tan B cot A

2 1

13 6

tan A

cos B sin A

13 7

tan B cot A

6 13 13 13

16 4 . Therefore:

3 sin A 5

tan A

13 . Therefore:

7 First draw the triangle:

Note that b

cos B sin A

13 7

sin B cos A

2 2 5 5 5

cos A

tan A

4 3 sin B cos A cos B sin A 5 5 y y Since y ≤ r, 1. Therefore sin 1, so it is impossible for sin 2 . r r

sin14 cos 9014 cos 76

Simplifying: sin 2 45 cos 2 60

Simplifying: tan 45 cot 45 11 2

2

1



2  2

1 1 3 2

3 4

tan B cot A

4 3

3

3 2

3 3 0 4 4

Simplifying: sin 2 60 cos 2 30

Simplifying:

10. 11.

Adding: 4831 2452 7283 7323 Converting to degrees and minutes: 73.2 73 0.2 73 0.2(60) 7312

12.

Converting to decimal degrees:

13.

Calculating the value:

14.

Calculating the value: cos 48.3 0.6652

15.

Calculating the value:

16.

18.

Since sin 0.6459 , sin 1 (0.6459 ) 40.2 . 1 1 Since sec 1.923 , cos , so cos1 1.923 First sketch the triangle:

19.

Using the Pythagorean Theorem: c 104 2 68 2 124 . Therefore: 68 tan A 104 68   A tan1 33.2 104  B 90 33.2 56.8 First sketch the triangle:

17.

2 

1 1 sin 30 csc 30 2

Using the Pythagorean Theorem: b 24.3 sin A 48.1 24.3  30.3 A sin1 48.1  B 90 30.3 59.7

58.7 . 1.923

48.12 24.32 41.5 . Therefore:

20.

First sketch the triangle:

Note that A = 90° – 24.9° = 65.1°. Therefore: tan 65.1

305 a 305 tan 65.1 657

21.

First sketch the triangle:

22.

Note that B 90 3530 8960 3530 5430 . Also: a sin 35.5 0.462 a 0.462 sin 35.5 0.268 First sketch the triangle:

Therefore: 52 x x sin 71 52 sin 71

52 55 cm sin 71

305 c c cos 65.1 305 cos 65.1

305 724 cos 65.1

b 0.462 b 0.462 cos 35.5 0.376

cos 35.5

23.

Sketch the figure:

Therefore: tan 75.5

24.

1.5 h 1.5 tan 75.5 5.8 The post is approximately 5.8 feet tall. Draw the figure:

Therefore: 35 x x tan 47 35 tan 47

35 32.64 feet tan 47 The stakes are 32.6 + 37.5 ≈ 70 feet apart. Let represent the required angle. Then: 31 tan 11  31 tan 1  70 11  The magnitudes are given by: Vx 850 cos 52 523 ft/sec 520 ft/sec x

25.

26.

Vy 850 sin 52 670 ft/sec

35 y y tan 43 35 tan 43

35 37.53 feet tan 43

27.

Draw the figure:

28.

Therefore the distances the ship has traveled are: east: 120 cos 30 100 miles Drawing the figure:

29.

Now find the magnitude of H: H tan 25.5 95.5 H 95.5 tan 25.5 45.6 Kelly must push horizontally with a force of 45.6 lb. The corresponding force diagram would be:

30.

Now find the magnitude of F: F sin 8.5 58.0 F 58.0 sin 8.5 8.57 lb Tyler must push with a force of 8.57 lb. The horizontal portion of the force is given by: Fx F cos 40 44 cos 40 lb

south: 120 sin 30 60 miles

The work is then given by: Work 44 cos 40 85 2, 865 ft-lb 2, 900 ft-lb ft-lb

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