Solutions for Trigonometry 11th Us Edition by Lial by 6alsm

Chapter 2

Acute Angles and Right Triangles

Section 2.1 Trigonometric Functions of Acute Angles

For Exercises 1 6, refer to the Function Values of Special Angles chart on page 52 of the text.

1. 1 C;sin30 2 

2. 2 H;cos45 2 

3. B;tan 451 

6. 3 2 1 2 cos3032 A;cot303 sin3021

7. side opposite21 sin hypotenuse29 A

side adjacent20 cos hypotenuse29 side opposite21 tan side adjacent20 A A

8. side opposite45 sin hypotenuse53 A

side adjacent28 cos hypotenuse53 side opposite45 tan side adjacent28 A A

9. side opposite sin hypotenuse An p  side adjacent cos hypotenuse side opposite tan side adjacent Am p An m

10. side opposite sin hypotenuse k A z  side adjacent cos hypotenuse side opposite tan side adjacent Ay z k A y 

11. a = 5, b = 12 2222222512169 13 cabcc c   side opposite12 sin hypotenuse13 side adjacent5 cos hypotenuse13 side opposite12 tan side adjacent5 b B c a B c b B a

  side adjacent5 cot side opposite12 hypotenuse13 sec side adjacent5 hypotenuse13 csc side opposite12 a B b c B a c B b

12. a = 3, b = 4

22222223425 5 cabcc c   side opposite4 sin hypotenuse5 side adjacent3 cos hypotenuse5 side opposite4 tan side adjacent3 b B c a B c b B a 

 side adjacent3 cot side opposite4 hypotenuse5 sec side adjacent3 hypotenuse5 csc side opposite4 a B b c B a c B b

13. a = 6, c = 7

222222 22 76 49361313 cabb bbb

side opposite13 sin hypotenuse7

side adjacent6 cos hypotenuse7

side opposite13 tan side adjacent6

 

side adjacent6 cot side opposite 13 613613 13 1313 hypotenuse7 sec side adjacent6 hypotenuse7

csc side opposite 13 713713 13 1313 a B

14. b = 7, c = 12 222222 22 127 144499595 caba aaa

side opposite7 sin hypotenuse12

side adjacent95 cos hypotenuse12 side opposite7 tan side adjacent 95 795795 95 9595

 side adjacent95 cot side opposite7 hypotenuse12 sec side adjacent 95 12951295 95 9595 hypotenuse12



csc side opposite7 a B b c B a c B b

15. a = 3, c = 10

222222 22 103 10099191 cabb bbb 



 side opposite91 sin hypotenuse10 side adjacent3 cos hypotenuse10 b B c a B c



side opposite91 tan side adjacent3 side adjacent3 cot side opposite 91 391391 91 9191 hypotenuse10 sec side adjacent3 hypotenuse10 csc side opposite 91 10911091 91 9191 b B a a B b c B a c B b



16. b = 8, c = 11 222222 22 118 121645757 caba aaa   side opposite8 sin hypotenuse11 side adjacent57 cos hypotenuse11 side opposite8 tan side adjacent 57 857857



57 5757 b B c a B c b B a



side adjacent57 cot side opposite8 hypotenuse11 sec side adjacent 57 11571157 57 5757 hypotenuse11 csc side opposite8 a B b c B a c B b







17. a = 1, c = 2 222222 22 21 4133 cabb bbb



 side opposite3 sin hypotenuse2 side adjacent1 cos hypotenuse2 side opposite3 tan3 side adjacent1 side adjacent1 cot side opposite 3 133 3 33 b B c a B c b B a a B b



 (continuedonnextpage)

Chapter 2 Acute Angles and Right Triangles

(continued)

hypotenuse2 sec2 side adjacent1 hypotenuse2 csc side opposite 3 2323 3 33 c B a c B b

18. 2,2ac 2 22222 22 22 4222 cabb bbb

side opposite2 sin hypotenuse2 side adjacent2 cos hypotenuse2 side opposite2 tan1 side adjacent 2 side adjacent2 cot1 side opposite 2 hypotenuse2 sec side adjacent 2 2222 2 2 22 hyp csc

otenuse2 side opposite 2 2222 2 2 22 c b

19. b = 2, c = 5 222222 22 52 2542121 caba aaa 

 side opposite2 sin hypotenuse5 side adjacent21 cos hypotenuse5 side opposite2 tan side adjacent 21 221221 21 2121 side adjacent21 cot side opposite2 hypotenuse5 sec side adjacent 21 5 21 b

 21521 21 21 hypotenuse5 csc side opposite2 c B b  

20.

21.

22.

23.

25.

26.

sincos90;cossin90; tancot90;cottan90; seccsc90;cscsec90

  

  

  cos30sin9030sin60 

  sin45cos9045cos45 

  csc60sec9060sec30 

24.   cot73tan9073tan17 

  sec39csc9039csc51 

  tan25.4cot9025.4cot64.6 

27.   sin38.7cos9038.7cos51.3 

28.

29.

cos20sin9020 sin70     



sec15csc9015 csc75  

30. Using 50,102,248, and 26,   we see that   sin90   and cos  yield the same values.

For exercises 31 40, if the functions in the equations are cofunctions, then the equations are true if the sum of the angles is 90º.

31.  

tancot10 1090 21090 28040 

32.   cossin230 23090 33090 312040        

33.

34.

35.

sin210cos320 21032090 51090 510020

sec10csc220 1022090 33090 36020

tan34cot510 3451090 8690 89612

36.

cot52tan24 522490 7690 78412

37.

38.

39.

sin20cos25 202590 31590 310535

cos250sin220 25022090 43090 46015 

sec310csc8 310890 41890 47218 

40. 



 csc40sec20 402090 22090 27035   

41. sin50sin40 

In the interval from 0º to 90º, as the angle increases, so does the sine of the angle, so sin 50º > sin 40º is true.

42. tan28tan40 

In the interval from 0º to 90º, as the angle increases, the tangent of the angle increases, so tan 40º > tan 28º tan28tan40  is true.

43. sin46cos46 

Using the cofunction identity,   cos46sin9046sin44  . In the interval from 0º to 90º, as the angle increases, so does the sine of the angle, so sin46sin44sin46cos46  is false.

44. cos28sin28 

Using the cofunction identity,   sin28cos9028cos62  . In the interval from 0º to 90º, as the angle increases, the cosine of the angle decreases, so cos28cos62cos28sin28  is false.

45. tan41cot41 

Using the cofunction identity,   cot41tan9041tan49  . In the interval from 0º to 90º, as the angle increases, the tangent of the angle increases, so tan41tan49tan41cot41  is true.

46. cot30tan40 

Using the cofunction identity,   cot30tan9030tan60  . In the interval from 0º to 90º, as the angle increases, the tangent of the angle increases, so tan60tan40cot30cot40  is false.

47. sec60sec30 

In the interval from 0º to 90º, as the angle increases, the cosine of the angle decreases, so the secant of the angle increases. Thus, sec60sec30  is true.

48. csc20csc30 

In the interval from 0º to 90º, as the angle increases, sine of the angle increases, so cosecant of the angle decreases. Thus csc20csc30  is false.

Use the following figures for exercises 49 64.

49. side opposite1 tan30 side adjacent 3 133 3 33 

50. side adjacent3 cot303 side opposite1 

51. side opposite1 sin30 hypotenuse2 

52. side adjacent3 cos30 hypotenuse2 

53. hypotenuse2 sec30 side adjacent 3 2323 3 33  

54. hypotenuse2 csc302 side opposite1 

55. hypotenuse2 csc452 side opposite1 

56. hypotenuse2 sec452 side adjacent1 

57. side adjacent1 cos45 hypotenuse 2 122 2 22

58. side adjacent1 cot451 side opposite1 

59. side opposite1 tan451 side adjacent1 

60. side opposite1 sin45 hypotenuse 2 122 2 22 

61. side opposite3 sin60 hypotenuse2 

62. side adjacent1 cos60 hypotenuse2 

63. side opposite3 tan603 side adjacent1 

64. hypotenuse2 csc60 side opposite 3 2323 3 33 



65. Because 3 sin60 2  and 60° is between 0° and 90°, A60. 

66. 0.7071067812 is a rational approximation for the exact value 2 2 (an irrational value).

67.

The line passes through (0, 0) and   3,1. The slope is change in y over the change in x

Thus, 1133 3 333 m  and the equation of the line is 3 3 yx 

68.

The line passes through   0,0 and   1,3.

The slope is change in y over the change in x

Thus, 3 3 1 m  and the equation of the line is 3.yx 

69. One point on the line 3 yx  is the origin   0,0. Let   , xy be any other point on this line. Then, by the definition of slope, 0 3, 0 yy m xx  but also, by the definition of tangent, tan3.   Because tan603,  the line 3 yx  makes a 60° angle with the positive x-axis (See exercise 68).

70. One point on the line 3 , 3 yx  is the origin   0,0. Let   , xy be any other point on this line. Then, by the definition of slope, 03 , 03 yy m xx  but also, by the definition of tangent, 3 tan. 3   Because 3 tan30, 3  the line 3 3 yx  makes a 30° angle with the positive x-axis. (See Exercise 67).

71. (a) Each of the angles of the equilateral triangle has measure   1 3 18060

(b) The perpendicular bisects the opposite side so the length of each side opposite each 30º angle is k.

The length of the perpendicular is 3k

(d) In a 30º-60º right triangle, the hypotenuse is always 2 times as long as the shorter leg, and the longer leg has a length that is 3 times as long as that of the shorter leg. Also, the shorter leg is opposite the 30º angle, and the longer leg is opposite the 60º angle.

72. (a) The diagonal forms two isosceles right triangles. Each angle formed by a side of the square and the diagonal measures 45º.

(b) By the Pythagorean theorem, 2222222 kkckcck 

Thus, the length of the diagonal is 2 k

73. Apply the relationships between the lengths of the sides of a 3060 right triangle first to the triangle on the left to find the values of y and x, and then to the triangle on the right to find the values of z and w. In the 3060 right triangle, the side opposite the 30 angle is 1 2 the length of the hypotenuse. The longer leg is 3 times the shorter leg.

Thus, we have  9 2 1993 9 and 3 222 9333 3, so, 62 33 33 and 2, so 233 2 yxy y yzz

74. Apply the relationships between the lengths of the sides of a 3060 right triangle first to the triangle on the left to find the values of a and b. In the 3060 right triangle, the side opposite the 30 angle is 1 2 the length of the hypotenuse. The longer leg is 3 times the shorter leg.

Apply the relationships between the lengths of the sides of a 4545 right triangle next to the triangle on the right to find the values of d and c. In the 4545 right triangle, the sides opposite the 45 angles measure the same.

The hypotenuse is 2 times the measure of a leg. 123 db and

21232126 cd

75. Apply the relationships between the lengths of the sides of a 4545 right triangle to the triangle on the left to find the values of p and r. In the 4545 right triangle, the sides opposite the 45 angles measure the same.

The hypotenuse is 2 times the measure of a leg.

Thus, we have 15 and 2152 prp

Apply the relationships between the lengths of the sides of a 3060 right triangle next to the triangle on the right to find the values of q and t. In the 3060 right triangle, the side opposite the 60 angle is 3 times as long as the side opposite to the 30 angle. The length of the hypotenuse is 2 times as long as the shorter leg (opposite the 30 angle).

Thus, we have 3 rq 1521523 56 3333 r q  and 

76. Apply the relationships between the lengths of the sides of a 3060 right triangle first to the triangle on the right to find the values of m and a. In the 3060 right triangle, the side opposite the 60 angle is 3 times as long as the side opposite to the 30 angle. The length of the hypotenuse is 2 times as long as the shorter leg (opposite the 30 angle).

Thus, we have 77373 73 and 3 333 mm  73143 22 33 ama

Apply the relationships between the lengths of the sides of a 4545 right triangle next to the triangle on the left to find the values of n and q. In the 4545 right triangle, the sides opposite the 45 angles measure the same. The hypotenuse is 2 times the measure of a leg. Thus, we have 143 3 na and 143146 22 33 qn

77. Because 1 , 2 Abh  we have 2 2 11 or . 222 AsssAs

78. Let h be the height of the equilateral triangle. h bisects the base, s, and forms two 30°–60° right triangles.

The formula for the area of a triangle is 1 . 2 Abh  In this triangle, b = s. The height h of the triangle is the side opposite the 60° angle in either 30°–60° right triangle. The side opposite the 30° angle is 2 s . The height is 3 3. 22 ss  So the area of the entire triangle is 2 133 224 Asss  

79. Graph the equations in the window [–1.5, 1.5] × [–1.2, 1.2]. The point of intersection is (0.70710678, 0.70710678). This corresponds to the point 22 , . 22

These coordinates are the sine and cosine of 45°.

80. Yes, the third angle can be found by subtracting the given acute angle from 90°, and the remaining two sides can be found using a trigonometric function involving the known angle and side.

2 sin454sin45422 42 y y  and 2 cos454cos45422 42 x x 

83. The legs of the right triangle provide the coordinates of P,   22,22.

84. 3 sin602sin6023 22 y y  and 1 cos602cos6021 22 x x 

The legs of the right triangle provide the coordinates of P. P is   1,3.

Section 2.2 Trigonometric Functions of Non-Acute Angles

1. The value of sin 240° is negative because 240° is in quadrant III. The reference angle is 60° , and the exact value of sin 240° is 3 2

2. The value of cos 390° is positive because 390° is in quadrant I. The reference angle is 30° , and the exact value of cos 390° is 3 2

3. The value of tan (–150°) is positive because –150° is in quadrant III. The reference angle is 30° and the exact value of tan (–150°) is 3 3

4. The value of sec 135° is negative because 135° is in quadrant II. The reference angle is 45° , and the exact value of sec 135° is 2.

5. C; 1809882  (98° is in quadrant II)

6. F; 212180 = 32°  (212° is in quadrant III)

7. A; 135360225  and 225180 = 45°  (225° is in quadrant III)

8. B; 60360300  and 36030060  (300° is in quadrant IV)

9. D; 750236030  (30° is in quadrant I)

10. B; 480360120  and 18012060  (120° is in quadrant II)

19. To find the reference angle for 300,  sketch this angle in standard position.

The reference angle is 36030060. 

Because 300° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. 3 sin300sin60 2 1 cos300cos60 2 tan300tan603 3 cot300cot60 3 sec300sec602 23 csc300csc60 3











20. To find the reference angle for 315,  sketch this angle in standard position.

The reference angle is 36031545.  Because 315° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.



2 sin315sin45 2 2 cos315cos45 2 tan315tan451 cot315cot451 sec315sec452 csc315csc452









21. To find the reference angle for 405°, sketch this angle in standard position.

The reference angle for 405° is 405° – 360° = 45°. Because 405° lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 45° See the Function Values of Special Angles table on page 52.)



2 sin405sin45 2 2 cos405cos45 2 tan405tan451 cot405cot451 sec405sec452 csc405csc452



22. To find the reference angle for 420º, sketch this angle in standard position.

The reference angle for 420º is 420º 360º = 60º. Because 420º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60°. See the Function Values of Special Angles table on page 52.)



   3 sin420sin60 2 1 cos420cos60 2 tan420tan603



 (continuedonnextpage)

(continued)

23. To find the reference angle for 480º, sketch this angle in standard position.

480º is coterminal with 480º 360º = 120º. The reference angle is 180º 120º = 60º. Because 480º lies in quadrant II, the cosine, tangent, cotangent, and secant are negative.





sec80sec602 23 csc480csc60 3

24. To find the reference angle for 495º, sketch this angle in standard position.

495º is coterminal with 495º 360º = 135º. The reference angle is 180º 135º = 45º. Because 495° lies in quadrant II, the cosine, tangent, cotangent, and secant are negative. 2 sin495sin45 2 2 cos495cos45 2 tan495tan451





cot495cot451 sec495sec452 csc495csc452





25. To find the reference angle for 570º sketch this angle in standard position.

570° is coterminal with 570º 360º = 210º. The reference angle is 210º 180º = 30º. Because 570º lies in quadrant III, the sine, cosine, secant, and cosecant are negative. 1 sin570sin30 2 3 cos570cos30 2 3 tan570tan30 3 cot570cot303 23 sec570sec30 3 csc570csc302







26. To find the reference angle for 750°, sketch this angle in standard position.

750° is coterminal with 30 because 750236075072030.  Because 750 lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 30.  1 sin750sin30 2 3 cos750cos30 2 3 tan750tan30 3



 (continuedonnextpage)

(continued)



cot750cot303 23 sec750sec30 3 csc750csc302



27. 1305º is coterminal with 1305336013051080225  . The reference angle is 225º 180º = 45º. Because 1305º lies in quadrant III, the sine, cosine, and secant and cosecant are negative. 2



sin1305sin45 2 2 cos1305cos45 2 tan1305tan451





The reference angle for 300  is 30036060.  Because 300  lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60°. See the Function Values of Special Angles table on page 52.)





cot1305cot451





sec1305sec452

      3 sin300sin60 2 1 cos300cos60 2 tan300tan603 3 cot300cot60 3 sec300sec602 23 csc300csc60 3



csc1305csc452

28. 1500º is coterminal with 150043601500144060  .

Because 1500º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60°.



   3 sin420sin60 2 1 cos420cos60 2 tan420tan603 3 cot420cot60 3 sec420sec602 23 csc420csc60 3



29. To find the reference angle for 300, sketch this angle in standard position.

30. 390° is coterminal with 3902360390720330.  The reference angle is 36033030.  Because 390° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.







  1 sin390sin30 2 3 cos390cos30 2 3 tan390tan30 3



    cot390cot303 23 sec390sec30 3 csc390csc302





31. –510° is coterminal with 5102360510720210.  The reference angle is 21018030.  Because –510° lies in quadrant III, the sine, cosine, and secant and cosecant are negative.

 



 1 sin510sin30 2 3 cos510cos30 2 3 tan510tan30 3





 



  cot510cot303 23 sec510sec30 3 csc510csc302



32. 1020º is coterminal with 102033601020108060  . Because 1020º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60. 



3 sin420sin60 2 1 cos420cos60 2 tan420tan603 3 cot420cot60 3 sec420sec602 23 csc420csc60 3



33. 1290° is coterminal with 1290436012901440150.  The reference angle is 18015030.  Because 1290° lies in quadrant II, the cosine, tangent, cotangent, and secant are negative. 1 sin2670sin30 2 3 cos2670cos30 2 3 tan2670tan30 3





 cot2670cot303 23 sec2670sec30 3 csc2670csc302





34. 855º is coterminal with 85533608551080225  . The reference angle is 225º 180º = 45º. Because 855º lies in quadrant III, the sine, cosine, and secant and cosecant are negative.



2 sin855sin45 2 2 cos855cos45 2 tan855tan451 cot855cot451 sec855sec452 csc855csc452







35. 1860º is coterminal with 1860636018602160300  . The reference angle is 360º 300º = 60º. Because 1860º lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.



      3 sin1860sin60 2 1 cos1860cos60 2 tan1860tan603 3 cot1860cot60 3 sec1860sec602 23 csc1860csc60 3









36. 2205º is coterminal with 2205736022052520315  The reference angle is 360º 315º = 45º. Because 2205º lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.





     2 sin2205sin45 2 2 cos2205cos45 2 tan2205tan451 cot2205cot451 sec2205sec452 csc2205csc452









37. Because 1305 is coterminal with an angle of 1305336013051080225,  it lies in quadrant III. Its reference angle is 22518045.  Because the sine is negative in quadrant III, we have 2 sin1305sin45. 2 

38. Because 1500 is coterminal with an angle of 150043601500144060,  it lies in quadrant I. Because 1500 lies in quadrant I, the values of all of its trigonometric functions will be positive, so 3 sin1500sin60. 2 

39. Because 510  is coterminal with an angle of 5102360510720210,  it lies in quadrant III. Its reference angle is 21018030.  The cosine is negative in quadrant III, so  3 cos510cos30. 2 

40. Because 1020  is coterminal with an angle of 102033601020108060,  it lies in quadrant I. Because 1020  lies in quadrant I, the values of all of its trigonometric functions will be positive, so

 tan1020tan603. 

41. Because 855º is coterminal with 85533608551080225  , it lies in quadrant III. Its reference angle is 225º 180º = 45º. The cosecant is negative in quadrant III, so   csc855csc452. 

42. Because 495º is coterminal with an angle of 4952360495720225  , it lies in quadrant III. Its reference angle is 22518045.  Tthe secant is negative in quadrant III, so   sec495sec452. 

43. Because 3015º is coterminal with 3015836030152880135  , it lies in quadrant II. Its reference angle is 180º 135º = 45º. The tangent is negative in quadrant II, so tan3015tan451. 

44. Because 2280º is coterminal with 2280636022802160120  , it lies in quadrant II. Its reference angle is 180º 120º = 60º. The cotangent is negative in quadrant II, so

50.

51.

2 2 cot135sin304tan45 119 14114 222

222 2 2 2 sin225cos270tan60 227 033 242

222 2 2 2 cot90sec180csc135 012121 

222 2 2 2 cos60sec150csc210 123 2 23 1429 4 4312

52.

244 4 2 4 cot135tan60sin180 1301910  

53.

 ? cos3060cos30cos60 

Evaluate each side to determine whether this statement is true or false.

  cos3060cos900  and 3131 cos30cos60 222   31 0, 2   so the statement is false.

54.  ? sin30sin60sin3060 

Evaluate each side to determine whether this statement is true or false.

1313 sin30sin60 222   and   sin3060sin901 

Because 13 1, 2   the given statement is false.

55. ? cos602cos30 

Evaluate each side to determine whether this statement is true or false. 1 cos60 2  and 3 2cos3023 2   

Because 1 3, 2  the statement is false.

56. ? 2 cos602cos301 

Evaluate each side to determine whether this statement is true or false. 1 cos60 2  and 2 2 33 2cos3012121 24 31 1 22

Because 11 , 22  the statement is true.

57. ? 22 sin45cos451  22 2222 1 2244

Because 1 = 1, the statement is true.

58. 22 tan601sec60 

Evaluate each side to determine whether this statement is true or false.

2 2 tan60131314  and 22 sec6024  . Because 4 = 4, the statement is true.

59.  ? cos2452cos45 

Evaluate each side to determine whether this statement is true or false.

  cos245cos900  and 2 2cos4522 2 

Because 02,  the statement is false.

60.  ? sin2302sin30cos30 

Evaluate each side to determine whether this statement is true or false.  3 sin230sin60 2 

2sin30cos302 222

Because 33 22  , the statement is true.

61. 1 sin 2  

Because sin  is positive,  must lie in quadrants I or II. Because one angle, namely 30,  lies in quadrant I, that angle is also the reference angle,  The angle in quadrant II will be 18018030150.   

62. 3 cos 2  

Because cos  is positive,  must lie in quadrants I or IV. One angle, namely 30,  lies in quadrant I, so that angle is also the reference angle,  The angle in quadrant IV will be 36036030330.   

63. tan3  

Because tan  is negative,  must lie in quadrants II or IV. The absolute value of tan  is 3, so the reference angle,  must be 60.  The quadrant II angle  equals 18018060120,    and the quadrant IV angle  equals 36036060300.   

64. sec2  

Because sec  is negative,  must lie in quadrants II or III. The absolute value of sec  is 2, so the reference angle,  must be 45.  The quadrant II angle  equals 18018045135,    and the quadrant III angle  equals 18018045225.   

65. 2 cos 2  

Because cos  is positive,  must lie in quadrants I or IV. One angle, namely 45,  lies in quadrant I, so that angle is also the reference angle,  The angle in quadrant IV will be 36036045315.   

66. 3 cot 3  

Because cot  is negative,  must lie in quadrants II or IV. The absolute value of cot  is 3 , 3 so the reference angle,  must be 60.  The quadrant II angle  equals 18018060120,    and the quadrant IV angle  equals 36036060300.   

67. csc2  

Because csc  is negative,  must lie in quadrants III or IV. The absolute value of csc  is 2, so the reference angle, ,  is 30 . The angle in quadrant III will be 18018030210    , and the quadrant IV angle is 36036030330   

68. 3 sin 2  

Because sin  is negative,  must lie in quadrants III or IV. The absolute value of sin  is 3 2 , so the reference angle, ,  is 60°. The angle in quadrant III will be 18018060240    , and the quadrant IV angle is 36036060300   

69. 3 tan 3  

Because tan  is positive,  must lie in quadrants I or III. One angle, namely 30 , lies in quadrant I, so that angle is also the reference angle, . The angle in quadrant III will be 18018030210   

70. 1 cos 2  

Because cos  is negative,  must lie in quadrants II or III. The absolute value of cos  is 1 , 2 so the reference angle,  must be 60.  The quadrant II angle  equals 18018060120,    and the quadrant III angle  equals 18018060240.   

71. csc2  

Because csc  is negative,  must lie in quadrants III or IV. The absolute value of csc  is 2, so the reference angle,  must be 45.  The quadrant III angle  equals 18018045225.    and the quadrant IV angle  equals 36036045315.   

72. cot1  

Because cot  is negative,  must lie in quadrants II or IV. The absolute value of cot  is 1, so the reference angle,  must be 45.  The quadrant II angle  equals 18018045135.    and the quadrant IV angle  equals 36036045315.   

73. 150º is in quadrant II, so the reference angle is 180º 150º = 30º.

3 cos30cos30633 2 x xr r  and 1 sin30sin3063 2 y yr r 

Because 150º is in quadrant II, the x- coordinate will be negative. The coordinates of P are  33,3

74. 225º is in quadrant III, so the reference angle is 225º 180º = 45º.

2 cos45cos451052 2 x xr r  and

2 sin45sin451052 2 y yr r 

Because 225º is in quadrant III, both the x- and y-coordinates will be negative. The coordinates of P are   52,52

75. For every angle  , 22 sincos1  . Because  22 0.80.60.640.361  , there is an angle  for which cos0.6   and sin0.8   . Because cos0   and sin0   ,  lies in quadrant IV.

76. For every angle  , 22 sincos1 

Because     22 39145 24 43169144 1  , there is no angle  for which 2 3 cos   and 3 4 sin   .

Chapter 2 Acute Angles and Right Triangles

77. If  is in the interval   90,180 , then 901804590 2    . Thus 2  lies in quadrant I, and cos 2  is positive.

78. If  is in the interval   90,180 , then 901804590 2    . Thus 2  lies in quadrant I, and sin 2  is positive.

79. If  is in the interval   90,180 , then 90180270180360   Thus 180   lies in quadrant IV, and   sec180   is positive.

80. If  is in the interval   90,180 , then 90180270180360  

Thus 180   lies in quadrant IV, and   cot180   is negative.

81. If  is in the interval   90,180 , then 9018090180 18090    

Because 180º is coterminal with 180º + 360º = 180º and 90º is coterminal with 90º + 360º = 270º,  lies in quadrant III, and   sin  is negative.

82. If  is in the interval   90,180 , then 9018090180 18090    

Because 180º is coterminal with 180º + 360º = 180º and 90º is coterminal with 90º + 360º = 270º,  lies in quadrant III, and   cos  is negative.

83. When an integer multiple of 360° is added to , the resulting angle is coterminal with  The sine values of coterminal angles are equal.

84. When an integer multiple of 360° is added to , the resulting angle is coterminal with . The cosine values of coterminal angles are equal.

85. The reference angle for 115° is 180° 115° = 65°. Because 115° is in quadrant II the cosine is negative. Sin  decreases on the interval (90°, 180°) from 1 to 0. Therefore, sin 115° is closest to 0.9.

86. The reference angle for 115° is 180° 115° = 65°. Because 115° is in quadrant II the cosine is negative. Cos  decreases on the interval (90°, 180°) from 0 to –1. Therefore, cos 115° is closest to 0.4.

87. When 45   , 2 sincos 2  . Sine and cosine are both positive in quadrant I and both negative in quadrant III. Because 18045180225,   45º is the quadrant I angle, and 225º is the quadrant III angle.

88. When 45   , 2 sincos 2  . Sine and cosine are opposites in quadrants II and IV. Thus, 18018045135   in quadrant II, and 36036045315   in quadrant IV.

Section 2.3 Approximations of Trigonometric Function Values

For Exercises 1 10, be sure your calculator is in degree mode.

1. J 2. B 3. E 4. F

5. D 6. I 7. H 8. A

9. G 10. C

For Exercises 11 40, be sure your calculator is in degree mode. If your calculator accepts angles in degrees, minutes, and seconds, it is not necessary to change angles to decimal degrees. Keystroke sequences may vary on the type and/or model of calculator being used. Screens shown will be from a TI-84 Plus C calculator. To obtain the degree (°) and (′) symbols, go to the ANGLE menu (2nd APPS).

For Exercises 11 40, we include TI-84 screens only for those exercises involving cotangent, secant, and cosecant.

11. sin38420.625243   12. cos41240.750111

sec13151.027349

csc145451.776815  

15. cot1834815.055723

16. tan421301.841771  

sin312120.740805

csc317361.483014  

20.   cot512201.907415  

21. 1 tan23.40.432739 cot23.4  

22. 1 cos14.80.966823 sec14.8  

23. cos77 cot770.230868 sin77   

24. sin33 tan330.649408 cos33   

25.   cot904.72tan4.720.082566 

26.   cos903.69sin3.690.064358 

27.  1 cos510.629320 csc9051  

28.

29.

30.

31.

32.

33.

 11 tan9022cot22 tan220.404026   

 1 tan1.4739716 tan1.473971655.845496    

 1 tan6.4358841 tan6.435884181.168073    

 1 sin0.27843196 sin0.2784319616.166641    

 1 sin0.84802194 sin0.8480219457.997172    

 11 cot1.2575516 1 cot1.2575516tan 1.2575516 38.491580

       

34.

csc1.3861147

csc1.3861147sin 1.3861147

35.

sec2.7496222

sec2.7496222cos

36.

sec1.1606249

sec1.1606249cos

37.

cos0.70058013

cos0.7005801345.526434

38.

cos0.85536428

cos0.8553642831.199998

39.

csc4.7216543sin 4.7216543

40.

cot0.21563481 1 cot0.21563481tan 0.21563481

41. A common mistake is to have the calculator in radian mode, when it should be in degree mode (and vice verse).

42. If the calculator allows an angle  where 0360   , then we need to find an angle within this interval that is coterminal with 2000º by subtracting a multiple of 360º: 2000536020001800200  . Then find cos 200°. If the calculator is more restrictive on evaluating angles (such as 090   ), then reference angles need to be used.

43. Find   1 1.482560 tan 969 A = 56º.

44. Find the sine of 22º. A = 0.3746065934º

45. sin35cos55cos35sin551 

46. cos100cos80sin100sin801 

47. 22 sin36cos361 

48. 2sin2513cos2513sin50260  

49. cos7529cos1431sin7529sin14310  

50. cos2814cos6146sin2814sin61461  

51. ? sin10sin10sin20  Using a calculator gives sin10sin100.34729636  and sin200.34202014.  Thus, the statement is false.

52. ? cos402cos20 

Using a calculator gives cos400.76604444  and 2cos201.87938524.  Thus, the statement is false.

53. ? sin502sin25cos25 

Using a calculator gives sin500.76604444  and 2sin25cos250.76604444.  Thus, the statement is true.

54. ? 2 cos702cos351 

Using a calculator gives cos700.34202014  and 2 2cos3510.34202014.  Thus, the statement is true.

55. ? 2 cos4012sin80 

Using a calculator gives cos400.76604444  and 2 12sin800.93969262.  Thus, the statement is false.

56. ? 2cos3822cos7644 

Using a calculator gives 2cos38221.56810939   and cos76440.22948353.   Thus, the statement is false.

57. ? sin3948cos39481  

Using a calculator gives sin3948cos39481.408393221.   Thus, the statement is false.

58.  ? 11 22 sin40sin40 

Using a calculator gives 1 2 sin400.32139380  and   1 2 sin400.34202014.  Thus, the statement is false.

59. ? 22 1cot42.5csc42.5 

Using a calculator gives 2 1cot42.52.1909542  and 2 csc42.52.1909542.  Thus, the statement is true.

60. ? 22 tan72251sec7225 

Using a calculator gives 2 tan7225110.9577102   and 2 sec722510.9577102.  

Thus, the statement is true.

61.  ? cos3020cos30cos20sin30sin20 

Using a calculator gives   cos30200.64278761  and cos30cos20sin30sin200.64278761. 

Thus, the statement is true.

62.  ? cos3020cos30cos20 

Using a calculator gives   cos30200.64278761  and cos30cos201.8057180.  Thus, the statement is false.

63. sin0.92718385   sin  is positive in quadrants I and II.

  1 sin0.9271838568

The angle in quadrant II with the same sine is 180° – 68° = 112°

64. sin0.52991926   sin  is positive in quadrants I and II.

  1 sin0.5299192632

The angle in quadrant II with the same sine is 180° – 32° = 148°

65. cos0.71933980   cos  is positive in quadrants I and IV.

  1 cos0.7193398044

The angle in quadrant IV with the same cosine is 360° – 44° = 316° .

66. cos0.10452846   cos  is positive in quadrants I and IV.

  1 cos0.1045284684

The angle in quadrant IV with the same cosine is 360° – 84° = 276°

67. tan1.2348971   tan  is positive in quadrants I and III.

  1 tan1.234897151

The angle in quadrant III with the same tangent is 180° + 51° = 231° .

68. tan0.70020753  

tan  is positive in quadrants I and III.

 1 tan0.7002075335

The angle in quadrant III with the same tangent is 180° + 35° = 215° .

69. sin 2100sin1.865.9670lb FW F

70.

F is negative because the car is traveling downhill.

71.

72.

73. sin 120 120sin(2.7) sin(2.7) 25472500lb

74. sin 145 145sin(3) sin(3) 27712800lb FW WW W

75. sin

2200sin276.77889275lb

2000sin2.276.77561818lb FW F F

The 2200-lb car on a 2° uphill grade has the greater grade resistance.

76.  sin  tan  180

(a) From the table, we see that if  is small, sintan 180   . (b) sintan 180 W FWW   

(d) Use 180 W F   from part (b). Let 3.75   and W = 1800. 1800(3.75) 118 lb 180 F  

77. 45 mph = 66 ft/sec, 66 ,3,32.2,Vg   0.14 f 

V R gf   

 22 66 tan32.2(0.14tan3) 703ft

78. There are 5280 ft in one mile and 3600 sec in one min. 1hr5280 ft 70 mph70mph· 3600 sec1 mi 2 102 ft per sec 3 102.67ft per sec    102.67 ,3,32.2,Vg   0.14 f    22 102.67 tan32.20.14tan3 1701ft V R gf   

79. Intuitively, increasing  would make it easier to negotiate the curve at a higher speed much like is done at a race track. Mathematically, a larger value of  (acute) will lead to a larger value for tan.  If tan  increases, then the ratio determining R will decrease. Thus, the radius can be smaller and the curve sharper if  is increased.

As predicted, both values are less.

80. From Exercises 77 amd 78,

Solving for V we have

ft/sec

80.9 ft/sec · 3600 sec/hr · 1 mi/5280 ft55 mph,  The speed limit should be 55 mph.

81. (a) 812146,31,310 c  m per sec



Because 1c is only given to one significant digit, 2c can only be given to one significant digit. The speed of light in the second medium is about 2108  m per sec.

(b) 812139,28,310 c  m per sec

  1112 2 221 8 8 2 sinsin sinsin 310sin28 210 sin39 cc c c c      

Because 1c is only given to one significant digit, 2c can only be given to one significant digit. The speed of light in the second medium is about 2108  m per sec.

82. (a) 81240,1.510 c   m per sec, and 8 1 310 c  m per sec     1121 2 221 8 2 8 8 1 2 8 sinsin sin sin 1.510sin40 sin 310 1.510sin40 sin19 310 cc cc 

(b) 81262,2.610 c   m per sec and 8 1 310 c  m per sec     1121 2 221 8 2 8 8 1 2 8 sinsin sin sin 2.610sin62 sin 310 2.610sin62 sin50 310 cc cc 

83. 81190,310 c   m per sec, and 8 2 2.25410 c     1121 2 221 8 2 8 8 8 1 2 sinsin sin sin 2.25410sin90 sin 310 2.254101 2.254 3 310 2.254 sin48.7 3 cc cc      

84.

Light from the object is refracted at an angle of 40.8° from the vertical. Light from the horizon is refracted at an angle of 48.7° from the vertical. Therefore, the fish thinks the object lies at an angle of 48.7° – 40.8° = 7.9° above the horizon.

85. 1 1hr5280 ft 55 mph55mph· 3600 sec1 mi 2 80 ft per sec80.67ft per sec, 3

3.5,   1 0.4, K  2 0.02. K

86. 1 80.67ft per sec, V  2 44 ft per sec, V  2,   12 0.4, and 0.02. KK

87. Negative values of  require greater distances for slowing down than positive values.

88. Using the values for 12 and KK from Exercise 73, determine 2V when D = 200, 3.5,   1 1hr5280 ft 90 mph90mph· 3600 sec1 mi 132 ft per sec

22 12 12 22 2 22 2 1.05 64.4sin 1.05132 200 64.40.40.02sin3.5 1.051321.05

23.12

13,671.21.05 13,671.2 1.05 13020.19048 114.106 V V V V V V

2 114ft/sec · 3600 sec/hr · 1 mi/5280 ft 78 mph

89. For Auto A, calculate 70cos1068.94. 

Auto A’s reading is approximately 69 mph. For Auto B, calculate 70cos2065.78. 

Auto B’s reading is approximately 66 mph.

90. The figure for this exercise indicates a right triangle. Because we are not considering the time involved in detecting the speed of the car, we will consider the speeds as sides of the right triangle. Given angle , cos. r a  

Thus, the speed that the radar detects is cos. ra   This confirms the “cosine effect” that reduces the radar reading.

91. h = 1.9 ft, 1 0.9,3,  2 4,   S = 336 ft:



 2 4(3)336 550 ft 2001.9336tan.9 L  

92. h = 1.9 ft, 1 1.5,3,  2 4,  

S = 336 ft:



 2 4(3)336 369 ft 2001.9336tan1.5 L  

Chapter 2 Quiz

(Sections 2.1 2.3)

1. side opposite243 sin hypotenuse405 A  side adjacent324 cos hypotenuse405 side opposite243 tan side adjacent324 A A

side adjacent324 cot side opposite243 hypotenuse405 sec side adjacent324 hypotenuse405 csc side opposite243

sin3036sin303618





4. The height of one of the six equilateral triangles from the solar cell is sinsin h hx x   .

Thus, the area of each of the triangles is 2 11 22 sin bhx    . So, the area of the solar cell is 22 1 2 6sin3sin xx   .

5. 180º 135º = 45º, so the reference angle is 45º. The original angle (135º) lies in quadrant II, so the sine and cosecant are positive, while the remaining trigonometric functions are negative.

2 sin135 2  ; 2 cos135 2 

tan1351  ; cot1351  sec1352  ; csc1352 

6. 150º is coterminal with 360º 150º = 210º. This lies in quadrant III, so the reference angle is 210º 180º = 30º. In quadrant III, the tangent and cotangent functions are positive, while the remaining trigonometric functions are negative.

  1 sin150sin30 2 3 cos150cos30 2

3 tan150tan30 3   



  cot150cot303 23 sec150sec30 3 csc150csc302



7. 1020º is coterminal with 1020º 720º = 300º. This lies in quadrant IV, so the reference angle is 360º 300º = 60º. In quadrant IV, the cosine and secant are positive, while the remaining trigonometric functions are negative. 3 sin1020sin60 2 1 cos1020cos60 2



 tan1020tan603 3 cot1020cot60 3 sec1020sec602 23 csc1020csc60 3







8. 3 sin 2  

Because sin  is positive,  must lie in quadrants I or II, and the reference angle, ,  is 60°. The angle in quadrant I is 60º, while the angle in quadrant II is 18018060120



 

9. sec2   Because sec  is negative,  must lie in quadrants II or III. The absolute value of sec  is 2, so the reference angle,  must be 45.  The quadrant II angle  equals 18018045135,    and the quadrant III angle  equals 18018045225.   

10. sin42180.673013  

11.   sec212121.181763  

12. tan2.674321069.497888 

13. csc2.386114724.777233 

14. The statement is false.

  sin6030sin901  , while 3131 sin60sin30 222   .

15. The statement is true. Using the cofunction identity,   tan9035cot35 

Section 2.4 Solutions and Applications of Right Triangles

1. B 2. D 3. A

4. F 5. C 6. E

7. 23.825 to 23.835

8. 28,999.5 to 29,000.5

9. 8958.5 to 8959.5

10. No. Points scored in basketball are exact numbers and cannot take on values that are not whole numbers.

11. If h is the actual height of a building and the height is measure as 58.6 ft, then 58.60.05. h 

12. If w is the actual weight of a car and the weight is measure as 1542 lb, then 15420.5. w 

13. 3620, A   c = 964 m

9090 903620 896036205340

ABBA B     sinsin3620 964 964sin3620571m (rounded to Aaa c a

    three significant digits)

coscos3620 964 964cos3620777 m (rounded to bb A c b

    three significant digits)

14. X = 47.8°, z = 89.6 cm

9090 9047.842.2

YXYX Y 

 sinsin47.8 89.6 89.6sin47.866.4cm xx X z x



 (rounded to three significant digits) coscos47.8 89.6 89.6cos47.860.2cm

Xyy z y



 (rounded to three significant digits)

15. N = 51.2°, m = 124 m

9090 9051.238.8

MNMN M   tantan51.2 124 124tan51.2154m (rounded to nn N m n 

 three significant digits) 124 coscos51.2 124



198m (rounded to cos51.2 m N pp p

  three significant digits)

16. 3140,35.9kmAa 

ABBA B 

 

9090 903140 896031405820

 35.9

sinsin3140

35.9 68.4 km (rounded to sin3140 Aa cc c  

  three significant digits) 35.9 tantan3140

 

35.9 58.2km (rounded to tan3140

  three significant digits)

17. 42.0892,56.851Bb

9090 9042.089247.9108

ABAB A   56.851

sinsin42.0892

b B cc c 

56.851 84.816 cm sin42.0892

  (rounded to five significant digits) 56.851 tantan42.0892

56.851 62.942 cm tan42.0892

b B aa a    (rounded to five significant digits)

18. 68.5142,3579.42Bc 9090 9068.514221.4858

ABAB A

sinsin68.5142 3579.42 3579.42sin68.51423330.68 m

B c b   (rounded to six significant digits) coscos68.5142 3579.42 3579.42cos68.51421311.04 m aa B c a   (rounded to six significant digits)

19. 12.5,16.2ab

Using the Pythagorean theorem, we have 222222 2 12.516.2 418.6920.5 ft (rounded to three abcc cc   significant digits)

 1 12.5 tantan 16.2 12.5 tan37.6540 16.2

370.65406037393740 a AA b A 

  (rounded to three significant digits)

 1 16.2 tantan 12.5 16.2 tan52.3460 12.5

520.3460605221 5220 (rounded to three significant digits) b BB a B      

20. 4.80,15.3ac Using the Pythagorean theorem, we have 222222 222 222 4.8015.3 4.8015.3

14.5 m (rounded to three abcb b b b  

  significant digits)

15.34.80211.05

 1 4.80 sinsin 15.3 4.80 sin18.2839 15.3 180.28396018171820 a AA b A 



  (rounded to three significant digits) (continuedonnextpage)

(continued)

(rounded to three significant digits)

21. No. You need to have at least one side to solve the triangle. There are infinitely many similar right triangles satisfying the given conditions.

22. If we are given an acute angle and a side in a right triangle, the unknown part of the triangle requiring the least work to find is the other acute angle. It may be found by subtracting the given acute angle from 90°.

23. Answers will vary. If you know one acute angle, the other acute angle may be found by subtracting the given acute angle from 90°. If you know one of the sides, then choose two of the trigonometric ratios involving sine, cosine or tangent that involve the known side in order to find the two unknown sides.

24. If you know the lengths of two sides, the Pythagorean theorem can be used to find the length of the remaining side. Then an inverse trigonometric function can be used to find one of the acute angles. The other acute angle may be found by subtracting the calculated acute angle from 90°.

25. A = 28.0°, c = 17.4 ft 90

AB BA B 

9028.062.0

sinsin28.0 17.4

17.4sin28.08.17ft Aaa c a 

 (rounded to three significant digits)

coscos28.00 17.4

17.4cos28.0015.4ft bb A c b



 (rounded to three significant digits)

26. B = 46.0°, c = 29.7 m

90 90 9046.044.0

AB AB A   

coscos46.0 29.7

29.7cos46.020.6m aa B c a



 (rounded to three significant digits)

sinsin46.0 29.7

B c b



29.7sin46.021.4m

 (rounded to three significant digits)

27. Solve the right triangle with B = 73.0°, b = 128 in. and C = 90° 9090 9073.017.0

ABAB A 

 128 tantan73.0



 

128 39.1in(rounded to tan73.0 three significant digits) 128 sinsin73.0



 

128 134in(rounded to sin73.0 three significant digits) b B aa aa b B cc cc

28. A = 62.5º, a = 12.7 m

9090 9062.527.5

Aa bb b



ABBA B   12.7 tantan62.5 12.7 6.61m(rounded to tan62.5 three significant digits)

  (continuedonnextpage)

(continued) 12.7

sinsin62.5

Aa cc c 

 

12.7 14.3m(rounded to sin62.5 three significant digits)

29. A = 61.0º, b = 39.2 cm

9090

ABBA B 

222222 22 2213 48416931518 m (rounded to two significant digits) cabb bbb



9061.029.0

Aaa b a

tantan61.0 39.2



39.2tan61.070.7cm

 (rounded to three significant digits)

39.2

coscos61.0

39.2 80.9cm cos61.0 b A cc c 

  (rounded to three significant digits)

30. B = 51.7º, a = 28.1 ft

9090

ABBB A 

We will determine the measurements of both A and B by using the sides of the right triangle. In practice, once you find one of the measurements, subtract it from 90 to find the other. 1 13 sinsin 22 13 sin36.221536 22 a AA c A 



(rounded to two significant digits) 1 13 coscos 22 13 cos53.778454 22 b BB c B

(rounded to two significant digits)

32. b = 32 ft, c = 51 ft

9051.738.3



tantan51.7 28.1





28.1tan51.735.6ft(rounded to three significant digits)

coscos51.7

bb B a b a B cc c

28.1



28.1 45.3ft (rounded to cos51.7 three significant digits)

31. a = 13 m, c = 22m

222222 22 5132 26011024157740 ft (rounded to two significant digits) caba aaa



(rounded to two significant digits) 1 32 sinsin 51 32 sin38.862339 51 b BB c B



 (rounded to two significant digits)

33. a = 76.4 yd, b = 39.3 yd

cabcab 

22222 22 76.439.35836.961544.49

7381.4585.9yd (rounded to three significant digits)

We will determine the measurements of both A and B by using the sides of the right triangle. In practice, once you find one of the measurements, subtract it from 90 to find the other.

(rounded to three significant digits)

34. a = 958 m, b = 489 m

2222222 958489

917,764239,1211,156,885 1075.5658871080 m (rounded to three significant digits)

We will determine the measurements of both A and B by using the sides of the right triangle. In practice, once you find one of the measurements, subtract it from 90 to find the other.

1 958 tantan 489 958 tan62.9585 489 630.95856062586300 a AA b A 

 (rounded to three significant digits)  1 489 tantan 958 489 tan27.0415 958 270.04156027022700 b BB a B 

(rounded to three significant digits)

35. a = 18.9 cm, c = 46.3 cm 222222 22 46.318.9 2143.69357.211786.48 42.3 cm (rounded to three significant digits)

cabb bb b

1 18.9 sinsin 46.3 18.9 sin24.09227 46.3 240.092276024062410 a AA c A 

(rounded to three significant digits)  1 18.9 coscos 46.3 18.9 cos65.9077 46.3 650.90776065546550 a BB c B 

(rounded to three significant digits)

36. b = 219 cm, c = 647 m 222 222 2 2 647219

418,60947,961 370,648 609 m cab a a a a      (rounded to three significant digits)

1 219 coscos 647 219 cos70.2154 647 700.21546070137010 b AA c A

(rounded to three significant digits)

1 219 sinsin 647 219 sin19.7846 647 190.78466019471950 b BB

(rounded to three significant digits)

37. A = 53º24′ , c = 387.1 ft 90 90 905324 89605324 3636 AB BA B 



sinsin5324



39. B = 39º09′ , c = 0.6231 m 90 90 903909 89603909 5051

AB BA B  

  

sinsin3909 0.6231

0.6231sin39090.3934m(rounded to four significant digits)

B c b     coscos3909 0.6231

B c a    

0.6231cos39090.4832m(rounded to four significant digits)

ABAB A 

  

40. B = 82º51′ , c = 4.825 cm 9090 90825189608251 709

sinsin8251 4.825

B c b    

4.825sin82514.787m(rounded to four significant digits)

coscos8251 4.825

B c a  

387.1

387.1sin5324310.8ft(rounded to four significant digits)

coscos5324

Aaa c a bb A c b

387.1

387.1cos5324230.8ft(rounded to four significant digits)

4.825cos82510.6006m(rounded to four significant digits)

41. If B is a point above point A as shown in the figure, the angle of elevation from A to B is the acute angle formed by the horizontal line through A and the line of sight from A to B.

ABBA B  

38. A = 13º47′ , c = 1285 m 9090 90134789601347 7613

 

42. No. An angle of elevation (and also an angle of depression) must be an acute angle.

Aaa c a bb

 

 

sinsin1347 1285 1285sin1347306.2m(rounded to four significant digits)

 

c b

 

coscos1347 1285 1285cos13471248m(rounded to four significant digits)

43. Angles DAB and ABC are alternate interior angle formed by the transversal AB intersecting parallel lines AD and BC. Thus they have the same measure.

44. An angle of depression (and also an angle of elevation) is measured from a horizontal line to the line of sight and. Angle CAB is formed by the line of sight and vertical line AC.

45. sin4350' 13.5 13.5sin4350'9.3496000

The ladder goes up the wall 9.35 m. (rounded to three significant digits)

46. T = 32° 10  and S = 5750

53.1m

Because 32 10 5750 896090, ST

triangle RST is a right triangle. Thus, we have

The distance across the lake is 33.4 m. (rounded to three significant digits)

47. Let x represent the horizontal distance between the two buildings and y represent the height of the portion of the building across the street that is higher than the window.

The height of the building across the street is about 128 ft. (rounded to three significant digits)

48. Let x = the diameter of the sun.

The included angle is 32,  so  1 3216. 2  

We will use this angle, d, and half of the diameter to set up the following equation.

The diameter of the sun is about 864,900 mi. (rounded to four significant digits)

49. The altitude of an isosceles triangle bisects the base as well as the angle opposite the base. The two right triangles formed have interior angles which have the same measure. The lengths of the corresponding sides also have the same measure. The altitude bisects the base, so each leg (base) of the right triangles is 42.36 2  21.18 in.

Let x = the length of each of the two equal sides of the isosceles triangle.

The length of each of the two equal sides of the triangle is 26.92 in. (rounded to four significant digits)

50. The altitude of an isosceles triangle bisects the base as well as the angle opposite the base. The two right triangles formed have interior angles which have the same measure. The lengths of the corresponding sides also have the same measure. The altitude bisects the base, so each leg (base) of the right triangles are 184.2 2  92.10 cm. Each angle opposite to the base of the right triangles measures

68443422.

Let h = the altitude.

In triangle ABC, 92.10 tan3422tan342292.10

The altitude of the triangle is 134.7 cm. (rounded to four significant digits)

51. Let h represent the height of the tower. In triangle ABC we have tan34.6

The height of the tower is 28.0 m. (rounded to three significant digits)

52. Let d = the distance from the top B of the building to the point on the ground A

In triangle ABC, 252 sin3230 252 469.0121 sin3230



The distance from the top of the building to the point on the ground is 469 m. (rounded to three significant digits)

53. Let x = the length of the shadow. 5.75 tan23.4

tan23.4 x x x

tan23.45.75 5.75 13.2875

The length of the shadow is 13.3 ft. (rounded to three significant digits)

54. Let x = the horizontal distance that the plan must fly to be directly over the tree.

10,500 tan1350 tan135010,500 10,500 42,641.2351 tan1350 x x x

The horizontal distance that the plan must fly to be directly over the tree is 42,600 ft. (rounded to three significant digits)

55. Let   the angle of depression.

1 39.8239.82 tantan 51.7451.74 37.583735

56. Let x = the height of the taller building; h = the difference in height between the shorter and taller buildings; d = the distance between the buildings along the ground.

(We hold on to these digits for the intermediate steps.) To find h, solve

Thus, the value of h rounded to three significant digits is 118 m. 28.011828.0146m,

so the height of the taller building is 146 m.

57. Let   the angle of elevation of the sun.

58. Let   the angle of elevation of the sun. 1 55.2055.20

59. In order to find the angle of elevation, , we need to first find the length of the diagonal of the square base. The diagonal forms two isosceles right triangles. Each angle formed by a side of the square and the diagonal measures 45˚

By the Pythagorean theorem, 22222 2 7007002700 27007002 dd dd

Thus, length of the diagonal is 7002 ft. To to find the angle, , we consider the following isosceles triangle.

The height of the pyramid bisects the base of this triangle and forms two right triangles. We can use one of these triangles to find the angle of elevation, 

Rounding this figure to two significant digits, we have 22.

60. Let y = the height of the spotlight (this measurement starts 6 feet above ground)

tan30.0 1000 1000tan30.0577.3502 y y 

Rounding this figure to three significant digits, we have 577. y 

However, the observer’s eye-height is 6 feet from the ground, so the cloud ceiling is 5776583  ft.

61. (a) Let x = the height of the peak above 14,545 ft.

The diagonal of the right triangle formed is in miles, so we must first convert this measurement to feet. Because there are 5280 ft in one mile, we have the length of the diagonal is   27.01345280  142,630.752. Find the value of x by solving sin5.82. 142,630.752 x 

142,630.752sin5.82 14,463.2674 x  

Thus, the value of x rounded to five significant digits is 14,463 ft. Thus, the total height is about 14,545 + 14,463 = 29,008 ≈ 29,000 ft.

(b) The curvature of the earth would make the peak appear shorter than it actually is. Initially the surveyors did not think Mt. Everest was the tallest peak in the Himalayas. It did not look like the tallest peak because it was farther away than the other large peaks.

62. Let x = the distance from the assigned target.

In triangle ABC, we have tan0030 234,000 234,000tan003034.0339 x x

The distance from the assigned target is 34.0 mi. (rounded to three significant digits)

Section 2.5 Further Applications of Right Triangles

11. ( 4, 0)

The bearing of the airplane measured in a clockwise direction from due north is 270°. The bearing can also be expressed as N 90° W, or S 90° W.

12. (5, 0)

The bearing of the airplane measured in a clockwise direction from due north is 90°. The bearing can also be expressed as N 90° E, or S 90° E.

13. (0, 4)

The bearing of the airplane measured in a clockwise direction from due north is 0°. The bearing can also be expressed as N 0° E or N 0º W.

14. (0, 2)

The bearing of the airplane measured in a clockwise direction from due north is 180°. The bearing can also be expressed as S 0° E or S 0º W.

15. ( 5, 5)

The bearing of the airplane measured in a clockwise direction from due north is 315°. The bearing can also be expressed as N 45° W.

16. ( 3, 3)

The bearing of the airplane measured in a clockwise direction from due north is 225°. The bearing can also be expressed as S 45° W.

17. (2, 2)

The bearing of the airplane measured in a clockwise direction from due north is 135°. The bearing can also be expressed as S 45° E.

18. (2, 2)

The bearing of the airplane measured in a clockwise direction from due north is 45°. The bearing can also be expressed as N 45° E.

19. Let x = the distance the plane is from its starting point. In the figure, the measure of angle ACB is

38180128385290°.

Therefore, triangle ACB is a right triangle.

(continuedonnextpage)

(continued)

Because , drt  the distance traveled in 1.5 hr is (1.5 hr)(110 mph) = 165 mi. The distance traveled in 1.3 hr is (1.3 hr)(110 mph) = 143 mi. Using the Pythagorean theorem, we have 2222 2 16514327,22520,449 47,674218.3438

The plane is 220 mi from its starting point. (rounded to two significant digits)

20. Let x = the distance from the starting point. In the figure, the measure of angle ACB is   2718011727 6390. 

Therefore, triangle ACB is a right triangle.

Applying the Pythagorean theorem, we have 2222 2 55140302519,600 22,62522,625150.4161

The distance of the end of the trip from the starting point is about 150 km. (rounded to two significant digits)

21. Let x = distance the ships are apart. In the figure, the measure of angle CAB is 1304090.  Therefore, triangle CAB is a right triangle.

Because , drt  the distance traveled by the first ship in 1.5 hr is (1.5 hr)(18 knots) = 27 nautical mi and the second ship is (1.5hr)(26 knots) = 39 nautical mi.

Applying the Pythagorean theorem, we have 2222 2 27397291521 2250225047.4342

The ships are 47 nautical mi apart (rounded to 2 significant digits).

22. Let x = distance the ships are apart. In the figure, the measure of angle BCA is 3603225290.  Therefore, triangle BCA is a right triangle.

Because , drt  the distance traveled by the first ship in 2.5 hr is (2.5 hr)(17 knots) = 42.5 nautical mi and the second ship is (2.5hr)(22 knots) = 55 nautical mi.

Applying the Pythagorean theorem, we have 222242.5554831.25 4831.2569.5072

The ships are 70 nautical mi apart (rounded to 2 significant digits).

23. Let b = the distance from dock A to the coral reef C.

In the figure, the measure of angle CAB is 9058223138,  and the measure of angle CBA is 328222705822.  Because 3138582290,   ABC is a right triangle.

cos 2587 cos31382587cos3138 2587 2203 ft b A b b b 



24. Let C = the location of the ship, and let c = the distance between the lighthouses. 18012943 1796012943 5017 mBAC

5017394390,   so we have a right triangle. Thus, 3742 sin3943sin39433742

3742 5856.1020 sin3943 c

The distance between the lighthouses is 5856 m (rounded to four significant digits).

25. Let x = distance between the two ships.

The angle between the bearings of the ships is   180°28°10 + 61°5090°.  The triangle formed is a right triangle. The distance traveled at 24.0 mph is (4 hr) (24.0 mph) = 96 mi. The distance traveled at 28.0 mph is (4 hr)(28.0 mph) = 112 mi. Applying the Pythagorean theorem we have 2222 2 96112921612,544 21,76021,760147.5127

The ships are 148 mi apart. (rounded to three significant digits)

26. Let C = the location of the transmitter; a = the distance of the transmitter from B.

The measure of angle CBA is 905340896053403620. 

The measure of angle CAB is 903620896036205340.  A + B = 90°, so C = 90°. Thus, we have sinsin5340 2.502.50 2.50sin53402.0140

The distance of the transmitter from B is 2.01 mi. (rounded to 3 significant digits)

27. Let b = the distance from A to C and let c = the distance from A to B

Because the bearing from A to B is N 84° E, the measure of angle ABD is 180° 84° = 96°. The bearing from B to C is 38°, so the measure of angle ABC = 180° (96° + 38°) = 46°. The bearing of A to C is 52°, so the measure of angle BAC is 180° (52° + 84°) = 44°. The measure of angle C is 180° (44° + 46°) = 90°, so triangle ABC is a right triangle. The distance from A to B, labeled c, is 2.4(250) = 600 miles. sin46 600 600sin46430 mi bb c b  

28. The information in the example gives 64,82,mDACmEAB  and 26. mGBC The sum of the measures of angles EAB and FBA is 180º because they are interior angles on the same side of a transversal.

(continuedonnextpage)

(continued)

So 1801808298mFBAmEAB    180648234 mCAB and   180982656 mABC . Thus, angle C is a right angle. It takes 1.8 hours at 350 mph to fly from A to B, so 1.8350630 mi ABc . To find the distance from B to C, use cos B cos630cos56350 mi 630 a ABCa 

29. Draw triangle WDG with W representing Winston-Salem, D representing Danville, and G representing Goldsboro. Name any point X on the line due south from D.

The bearing from W to D is 42° (equivalent to N 42 E), so angle WDX measures 42°. Because angle XDG measures 48°, the measure of angle D is 42° + 48° = 90°. Thus, triangle WDG is a right triangle. Using d = rt and the Pythagorean theorem, we have

The distance from Winston-Salem to Goldsboro is approximately 140 mi. (rounded to two significant digits)

30. Let x = the distance from Atlanta to Augusta.

The line from Atlanta to Macon makes an angle of 27° + 63°90°,  with the line from Macon to Augusta. Because , drt  the distance from Atlanta to Macon is

 1 4 60175mi.  The distance from Macon to Augusta is   3 4 601105mi. 

Use the Pythagorean theorem to find x: 2222

75105562511,025 16,650129.0349

The distance from Atlanta to Augusta is 130 mi. (rounded to two significant digits)

31. Let x = the distance from the closer point on the ground to the base of height h of the pyramid.

In the larger right triangle, we have

tan2110'135tan2110' 135 h hx x

In the smaller right triangle, we have tan3530'tan3530'. h hx x 

Substitute for h in this equation, and solve for x to obtain the following.

 

135tan2110'tan3530' 135tan2110'tan2110'tan3530' 135tan2110'tan3530'tan2110' 135tan2110'tan3530'tan2110' 135tan2110' tan3530'tan2110' xx xx xx x x       

Substitute for x in the equation for the smaller triangle.

135tan2110' tan3530 tan3530'tan2110' 114.3427 h    

The height of the pyramid is 114 ft. (rounded to three significant digits)

32. Let x = the distance traveled by the whale as it approaches the tower; y = the distance from the tower to the whale as it turns.

68.7 tan354068.7tan3540 68.768.7andtan1550 tan3540 y y y xy 

The whale traveled 147 m as it approached the lighthouse. (rounded to three significant digits)

33. Let x = the height of the antenna; h = the height of the house.

In the smaller right triangle, we have tan181028tan1810. 28 h h 

In the larger right triangle, we have tan271028tan2710 28 28tan2710 28tan271028tan1810 5.1816 xh xh xh x   

The height of the antenna is 5.18 m. (rounded to three significant digits)

34. Let x = the height of Mt. Whitney above the level of the road; y = the distance shown in the figure below.

In triangle ADC, tan2240'tan2240' .(1) tan2240' x yx y x y 

In triangle ABC  tan1050' 7.00

7.00tan1050' tan1050'7.00tan1050' 7.00tan1050' (2) tan1050' x y yx yx x y       

Setting equations 1 and 2 equal, we have

7.00tan1050

tan2240tan1050 tan1050tan2240

7.00tan1050tan2240

7.00tan1050tan2240 tan2240tan1050

7.00tan1050tan2240 tan2240'tan1050' 7.00tan1050tan2240 tan2240

The height of the top of Mt. Whitney above road level is 2.47 km. (rounded to three significant digits)

35. Algebraic solution: Let x = the side adjacent to 49.2º in the smaller triangle.

In the larger right triangle, we have  tan29.5392tan29.5 392 h hx x   .

In the smaller right triangle, we have tan49.2tan49.2 h hx x  Substituting, we have    tan49.2392tan29.5 tan49.2392tan29.5 tan29.5 tan49.2tan29.5392tan29.5 tan49.2tan29.5392tan29.5 392tan29.5 tan49.2tan29.5 xx x x xx x x        

Now substitute this expression for x in the equation for the smaller triangle to obtain tan49.2

392tan29.5 tan49.2 tan49.2tan29.5 433.4762433 ft (rounded to three hx h      significant digits.

(continuedonnextpage)

(continued)

Graphing calculator solution:

Graph   1 tan29.5 yx  and     2 tan29.5392yx in the window [0, 1000] × [0, 500]. Then find the intersection.

The height of the triangle is 433 ft (rounded to three significant digits.

36. Algebraic solution:

Let x = the side adjacent to 52.5º in the smaller triangle.

Graphing calculator solution: Graph   1 tan41.2 yx  and

2 tan52.5168yx in the window [0, 800] × [0, 600]. Then find the intersection.

The height of the triangle is 448 m (rounded to three significant digits.

37. Let x = the minimum distance that a plant needing full sun can be placed from the fence.

4.65 tan2320tan23204.65 4.65 10.7799 tan2320 x x x  

The minimum distance is 10.8 ft. (rounded to three significant digits)

In the larger right triangle, we have 

38.

tan41.2168tan41.2 168 h hx x   .

In the smaller right triangle, we have

tan52.5tan52.5. h hx x 

Substituting, we have

tan52.5168tan41.2

tan41.2

tan52.5tan41.2168tan41.2 tan52.5tan41.2168tan41.2 168tan41.2 tan52.5tan41.2 xx x x xx x x

Now substitute this expression for x in the equation for the smaller triangle to obtain tan52.5

168tan41.2

tan52.5 tan52.5tan41.2 448.0432448 m (rounded to three hx h

significant digits.

1.0837 tan0.7261944649 1.4923 tan0.726194464935.987 3559.2355910 1.4923 tan1.377041617 1.0837

tan1.37704161754.013 5400.8540050

39. Let h = the minimum height above the surface of Earth so a pilot at A can see an object on the horizon at C

Using the Pythagorean theorem, we have

40004000125

400016,000,00015,625

400016,015,625

400016,015,62516,015,62540004001.952640001.9526

The minimum height above the surface of Earth would be 1.95 mi. (rounded to 3 significant digits)

40. Let y = the common hypotenuse of the two right triangles.

To find x, first find the angle opposite x in the right triangle: 52203050518030502130 

sin2130sin2130231.0571948sin213084.6827 231.0571948 xx x y

The length x is approximate 84.7 m. (rounded) 41. (a) cotcotcotcot 2222222

(b) Using the result of part (a), let 3748,4203,  and b = 2.000 cotcot 222 2.00037484203 cotcotcot0.315cot0.3504166667345.3951 222 b d d    

The distance between the two points P and Q is about 345.4 cm.

42.

2 2 sincoscossin64 32 vvvh D  

All answers are rounded to four significant digits.

(a) 44ft per sec and 7ft, vh  so

2 2 44sincos44cos44sin647 32 D

2 1936sin40cos4044cos4044sin40448

If 40,  

ft. 32 D  

2 1936sin42cos4244cos4244sin42448

If 42,  

If 45,  

2 1936sin45cos4544cos4544sin45448

32 D

As  increases, D increases and then decreases.

(b) 7ftand 42, h   so

2 2 sin42cos42cos42sin42647 32 vvv D  

2 2 43sin42cos4243cos4243sin42448

If 43, v 

32 D

2 2 44sin42cos4244cos4244sin42448

If 44, v 

ft 32 D

If 45, v 

2 2 45sin42cos4245cos4245sin42448 69.93 ft 32 D

As v increases, D increases.

(c) The velocity affects the distance more. The shot-putter should concentrate on achieving as large a value of v as possible.

43. (a) If 37,   then 37 18.5. 22   

To find the distance between P and Q, d, we first note that angle QPC is a right angle. Hence, triangle QPC is a right triangle and we can solve tan18.5 965 965tan18.5322.8845 d d   The distance between P and Q, is 320 ft. (rounded to two significant digits)

(b) We are dealing with a circle, so the distance between M and C is R. If we let x be the distance from N to M, then the distance from C to N will be Rx

Triangle CNP is a right triangle, so we can set up the following equation. coscos 22 cos1cos 22 Rx RRx R xRRxR 

44. (a)

(b)

The distance is 23 ft. (rounded to two significant digits)

45.

46.

47.

The distance is 48 ft. (rounded to two significant digits)

tantan3525yxayx

The line that bisects quadrants I and III passes through the origin, so a = 0. In addition, 45   because the line bisects the angle formed by the axes. tan451,  so an equation of the line is   tan45 yx  or y = x

48.

The line that bisects quadrants II and IV passes through the origin, so a = 0. In addition, 135   because the line bisects the angle formed by the axes and the reference angle  is 45° tan1351,  so an equation of the line is   tan135 yx  or y = –x

49. All points whose bearing from the origin is 240º lie in quadrant III.

The reference angle,  , is 30º. For any point, (x, y) on the ray cos x r    and sin, y r    where r is the distance from the point to the origin. Let r = 2, so 3 2 cos cos2cos3023 x r xr      1 2 sin sin2sin3021 y r yr       Thus, a point on the ray is  3,1 . The ray contains the origin, the equation is of the form y = mx. Substituting the point  3,1 , we have   13 m 

33 11 3 333 m  . Thus, an equation of the ray is 3 3 ,0yxx (because the ray lies in quadrant III).

50. All points whose bearing from the origin is 150º lie in quadrant IV.

The reference angle,  , is 60º. For any point, (x, y) on the ray cos x r    and sin, y r    where r is the distance from the point to the origin. Let r = 2, so

1 2 coscos2cos6021 x xr r

continuedonnextpage

Thus, a point on the ray is   1,3

The ray contains the origin, so the equation is of the form y = mx. Substituting the point

1,3 , we have 

Thus, an equation of the ray is 3,0yxx (because the ray lies in quadrant IV).

Chapter 2 Review Exercises

1. side opposite60 sin hypotenuse61 side adjacent11

cos hypotenuse61 side opposite60

tan side adjacent11

side adjacent11

cot side opposite60 hypotenuse61

sec side adjacent11 hypotenuse61

csc side opposite60

2. side opposite4020 sin hypotenuse5829 side adjacent4221 cos hypotenuse5829 side opposite4020 tan side adjacent4221

side adjacent4221

4.     sec210csc420 

Secant and cosecant are cofunctions, so the sum of the angles is 90º.

    21042090 63090 66010      

5.     tan511cot62 xx 

Tangent and cotangent are cofunctions, so the sum of the angles is 90º.

  5116290 111390 11777 xx x xx 

6. 37 cos11sin40 510  

Sine and cosine are cofunctions, so the sum of the angles is 90º. 37 114090 510 67 114090 1010 1313 519039 1010 10 3930 13

7. sin46sin58 

In the interval from 0º to 90º, as the angle increases, so does the sine of the angle, so sin 46º < sin 58º is true.

8. cos47cos58 

In the interval from 0º to 90º, as the angle increases, the cosine of the angle decreases, so cos47cos58  is false.

9. tan60cot40 

cot side opposite4020 hypotenuse5829 sec side adjacent4221 hypotenuse5829

csc side opposite4020 A A A

3. sin4cos5   

Sine and cosine are cofunctions, so the sum of the angles is 90º.

Using the cofunction identity,   cot40tan9040tan50  . In quadrant I, the tangent function is increasing. Thus cot40tan50tan60,  and the statement is true.

10. csc22csc68 

In quadrant I, the cosecant function is decreasing. Thus csc22csc68,  and the statement is false.

11.

cos b c A  and sin, b c B  so cossin. AB 

This is an example of equality of cofunctions of complementary angles.



   

 

 

12. If 135,18013545. If 45,45. If 300,36030060. If 140,18014040.



 

Of these reference angles, 40° is the only one which is not a special angle, so D, tan 140°, is the only one which cannot be determined exactly using the methods of this chapter.

13. 1020º is coterminal with 10202360300  . The reference angle is 36030060.  Because 1020° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.



3 sin1020sin60 2 1 cos1020cos60 2 tan1020tan603 3 cot1020cot60 3 sec1020sec602 23 csc1020csc60 3









14. A 120º angle lies in quadrant II, so the reference angle is 18012060.  Because 120° is in quadrant II, the cosine, tangent, cotangent, and secant are negative.



3 sin120sin60 2 1 cos120cos60 2



 tan120tan603 13 cot120cot60 3 3 sec120sec602 223 csc120csc60 3 3







15. –1470° is coterminal with 14705360330.  This angle lies in quadrant IV. The reference angle is 36033030.  Because –1470° is in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. 1 sin(1470)sin30 2 3 cos(1470)cos30 2 3 tan(1470)tan30 3



 cot(1470)cot303 23 sec(1470)sec30 3 csc(1470)csc302





16. –225° is coterminal with –225° + 360° = 135°. This angle lies in quadrant II. The reference angle is 18013545.  Because –225° is in quadrant II, the cosine, tangent, cotangent, and secant are negative. 2 sin(225)sin45 2 2 cos(225)cos45 2 tan(225)tan451





 cot(225)cot451 sec(225)sec452 csc(225)csc452





17. 1 cos 2  

18. 1 sin 2  

Because sin  is negative,  must lie in quadrants III or IV. The absolute value of sin  is 1 2 , so the reference angle, ,  is 30º.

The angle in quadrant III will be 18018030210    , and the quadrant IV angle is 36036030330    .

19. 23 sec 3  

Because sec  is negative,  must lie in quadrants II or III. The absolute value of sec  is 23 , 3 so the reference angle,  must be 30º. The quadrant II angle  equals 18018030150,    and the quadrant III angle  equals 18018030210.   

20. cot1  

21.

24. (a) ( 3, 3)

 33 sin 22 y r

1 cos 2 x r   3 tan3 1 y x 

For the exercises in this section, be sure your calculator is in degree mode.

25. 1 sec222301.356342 cos22230   

26. sin72300.953717  

27. 1 csc78211.021034 sin7821   

28. 1 cot305.60.715930 tan305.6  

29. tan11.76890.208344 

30. 1 sec58.90411.936213 cos58.9041  

31.  1 sin0.8254121 sin0.825412155.673870    

32.

 11 cot1.1249386 1 cot1.1249386tan 1.1249386 41.635092   





33.

 1 cos0.97540415 cos0.975404112.7353938    

34.

  



 11 sec1.2637891 1 sec1.2637891cos 1.2637891

37.695528



47. 39.72, 38.97 m Ab

49. Let x = the height of the tree.

ABBA B 

9090 9039.7250.28

tantan39.72 38.97 38.97tan39.7232.38 m(rounded to four significant digits) Aaa b

tan7050tan70137 ft 50 x x

50. 22 1 1251442516913 55 tantan23 1212 r

51. Let x = height of the tower.

(rounded to five significant digits)

48. 4753, 298.6 m Bb

9090 90475389604753 4207

tan4753298.6 298.6 270.0 m tan4753 b B aa a a

(rounded to four significant digits)

sin4753298.6

(rounded to four significant digits)

tan3820 93.2 93.2tan3820 73.6930 x x x

The height of the tower is 73.7 ft. (rounded to three significant digits)

52. Let h = height of the tower.

tan29.5 36.0 36.0tan29.520.3678 h h  

The height of the tower is 20.4 m. (rounded to three significant digits)

53. Let x = length of the diagonal 15.24 cos35.65 15.24 18.7548

The length of the diagonal is 18.75 cm (rounded to three significant digits).

54. Let x = the length of the equal sides of an isosceles triangle. Divide the isosceles triangle into two congruent right triangles

Each side is 50.24 m long (rounded to 4 significant digits).

55. Draw triangle ABC and extend the north-south lines to a point X south of A and S to a point Y, north of C

Angle ACB = 344° – 254° = 90°, so ABC is a right triangle.

Angle BAX = 32° because it is an alternate interior angle to 32°.

Angle YCA = 360° – 344° = 16°

Angle XAC = 16° because it is an alternate interior angle to angle YCA

Angle BAC = 32° + 16° = 48°.

In triangle ABC, 780 coscos48 780 cos487801165.6917

The distance from A to B is 1200 m. (rounded to two significant digits)

56. Draw triangle ABC and extend north-south lines from points A and B. Angle ABX is 55° (alternate interior angles of parallel lines cut by a transversal have the same measure) so Angle ABC is 55° + 35° = 90°.

Angle ABC is a right angle, so use the Pythagorean theorem to find the distance from A to C.

22 22 2 817465615476 12,03712,037109.7133 ACAC ACAC

It is 110 km from A to C. (rounded to two significant digits)

57. Suppose A is the car heading south at 55 mph, B is the car heading west, and point C is the intersection from which they start. After two hours, using d = rt, AC = 55(2) = 110. Angle ACB is a right angle, so triangle ACB is a right triangle. The bearing of A from B is 324º, so angle CAB = 360º 324º = 36º.

(continuedonnextpage)

(continued)

The distance from A to B is about 140 mi (rounded to two significant digits).

58. Let x = the leg opposite angle A tantanAxxkA k  and tantan hx BxkBh k   . So,

59. Answers will vary. Sample answer: Find the value of x.

60. Answers will vary. Sample answer: Find the value of x

61.

180 1 1 cos T P hR

(a) Let R = 3955 mi, T = 25 min, P = 140 min.

The height of the satellite is approximately 716 mi.

(b) Let R = 3955 mi, T = 30 min, P = 140 min.

The height of the satellite is approximately 1104 mi.

62. (a) From the figure, we see that sinsin. QP QP xx xxd d  

Similarly, we have coscos. QP QP yyyyd d  

(b) Let    ,123.62,337.95PPxy  , 171922,   and d = 193.86 ft. sin 123.62193.86sin171922 181.3427

QP Q xxd x 

QP Q yyd y   

  cos 337.95193.86cos171922 523.0170

Thus, the coordinates of Q are (181.34, 523.02), rounded to five significant digits.

Chapter 2 Chapter Test

1. side opposite12 sin hypotenuse13 side adjacent5 cos hypotenuse13 side opposite12 tan side adjacent5

side adjacent5 cot side opposite12 hypotenuse13 sec side adjacent5 hypotenuse13 csc side opposite12

2. Apply the relationships between the lengths of the sides of a 3060 right triangle first to the triangle on the right to find the values of y and w. In the 3060 right triangle, the side opposite the 60 angle is 3 times as long as the side opposite to the 30 angle. The length of the hypotenuse is 2 times as long as the shorter leg (opposite the 30 angle).

Thus, we have 43 y  and   248 w 

Apply the relationships between the lengths of the sides of a 4545 right triangle next to the triangle on the left to find the values of x and z. In the 4545 right triangle, the sides opposite the 45 angles measure the same. The hypotenuse is 2 times the measure of a leg. Thus, we have 4and 42 xz

3. 

 sin15cos230 

Sine and cosine are cofunctions, so the sum of the angles is 90º. So,

1523090

4. (a) sin24sin48 

In the interval from 0º to 90º, as the angle increases, so does the sine of the angle, so sin 24º < sin 48º is true.

(b) cos24cos48 

In the interval from 0º to 90º, as the angle increases, so the cosine of the angle decreases, so cos 24º < cos 48º is false.

cos6030cos900 cos60cos30sin60sin30 1331

Thus, the statement is true.

5. A 240º angle lies in quadrant III, so the reference angle is 24018060.  Because 240° is in quadrant III, the sine, cosine, secant, and cosecant are negative. 3 sin240sin60 2 1 cos240cos60 2 tan240tan603



 13 cot240cot60 3 3 sec240sec602 223 csc240csc60 3 3







6. –135° is coterminal with –135° + 360° = 225°. This angle lies in quadrant III. The reference angle is 22518045.  Because –135° is in quadrant III, the sine, cosine, secant, and cosecant are negative.



2 sin(135)sin45 2 2 cos(35)cos45 2 tan(135)tan451



 cot(135)cot451 sec(135)sec452 csc(135)csc452





7. 990º is coterminal with 9902360270  , which is the reference angle. sin990sin2701 cos990cos2700 tan990tan270 undefined





 cot990cot2700 sec990sec270undefined csc990csc2701







8. 2 cos 2  

Because cos  is negative,  must lie in quadrant II or quadrant III. The absolute value of cos  is 2 2 , so 45    . The quadrant II angle  equals 18018045135,    and the quadrant III angle  equals 18018045225.   

9. 23 csc 3  

Because csc  is negative,  must lie in quadrant III or quadrant IV. The absolute value of csc  is 23 3 , so 60    . The quadrant III angle  equals 18018060240,    and the quadrant IV angle  equals 36036060300.   

10. tan145 or 225 

11. tan1.6778490  

1 1 cottan, tan    so we can use division or the inverse key (multiplicative inverse).

12. (a) sin78210.979399   (b) tan117.6891.905608

13.  1 sin0.27843196 sin0.2784319616.166641

14. 5830,748Aa

9090 9058303130

748 tantan5830

748 458 (rounded to three tan5830 significant digits 748 sinsin5830 748 877 (rounded to three sin5830 significant digits

15. Let  = the measure of the angle that the guy wire makes with the ground.

16. Let x = the height of the flagpole.

The flagpole is approximately 15.5 ft high. (rounded to three significant digits)

17. Let h = the height of the top of mountain above the cabin. Then 2000 + h = the height of the mountain.

tan266800 14,000 h h  (rounded to two significant digits). Thus, the height of the mountain is about 6800 + 2000 = 8800 ft.

18. Let x = distance the ships are apart. In the figure, the measure of angle CAB is 1223290.  Therefore, triangle CAB is a right triangle.

Because d = rt, the distance traveled by the first ship in 2.5 hr is (2.5 hr)(16 knots) = 40 nautical mi and the second ship is (2.5hr)(24 knots) = 60 nautical mi. Applying the Pythagorean theorem, we have 2222 2 406016003600 5200520072.111 xx xx

The ships are 72 nautical mi apart (rounded to 2 significant digits).

19. Draw triangle ACB and extend north-south lines from points A and C. Angle ACD is 62° (alternate interior angles of parallel lines cut by a transversal have the same measure), so Angle ACB is 62° + 28° = 90°.

Angle ACB is a right angle, so use the Pythagorean theorem to find the distance from A to B

22 22 2 755356252809 8434843491.8368

It is 92 km from the pier to the boat, rounded to two significant digits.

20. Let x = the side adjacent to 52.5° in the smaller triangle.

In the larger triangle, we have  tan41.2168tan41.2. 168 h hx x  

In the smaller triangle, we have tan52.5tan52.5. h hx x 

Substitute for h in this equation to solve for x

tan52.5tan41.2 xx xx xx x x

168tan41.2tan52.5

168tan41.2tan41.2tan52.5 168tan41.2tan52.5tan41.2 168tan41.2tan52.5tan41.2 168tan41.2

Substituting for x in the equation for the smaller triangle gives tan52.5

tan52.5tan41.2 hx h

168tan41.2tan52.5 448.0432

The height of the triangle is approximately 448 m (rounded to three significant digits).

Chapter 2 Acute Angles and Right Triangles

Section 2.1 Trigonometric Functions of Acute Angles

Classroom Example 1 (page 48)

side opposite36 sin hypotenuse85 A  side adjacent77 cos hypotenuse85 side opposite36 tan side adjacent77

A A

side opposite77 sin hypotenuse85

B  side adjacent36 cos hypotenuse85 side opposite77 tan side adjacent36

B B 

Classroom Example 2 (page 49)

(a)   sin9cos909cos81 

(b)

(c)

 cot76tan9076tan14 

csc45sec9045sec45  

Classroom Example 3 (page 50)

(a)

 cot8tan413 

Cotangent and tangent are cofunctions, so the equation is true if the sum of the angles is 90º. 



   cot8tan413 8413905590 58517   

 

(b)     sec514csc28 

Secant and cosecant are cofunctions, so the equation is true if the sum of the angles is 90º.

sec514csc28 51428907690 78412

Classroom Example 4 (page 51)

(a) tan25tan23 

In the interval from 0º to 90º, as the angle increases, the tangent of the angle increases. Thus tan25tan23  is false.

(b) csc44csc40 

In the interval from 0º to 90º, as the angle increases, the sine of the angle increases, so the cosecant of the angle decreases. Thus, csc44csc40  is true.

Classroom Example 5 (page 52)

side opposite1 sin30 hypotenuse2  side adjacent3 cos30 hypotenuse2 side opposite1133 tan30 side adjacent3 333



 side adjacent3 cot303 side opposite1 hypotenuse22323 sec30 side adjacent3 333



 hypotenuse2 csc302 side opposite1 

Section

2.2

Trigonometric Functions of Non-Acute Angles

Classroom Example 1 (page 57)

(a) 294º lies in quadrant IV. The reference angle is 36029466  .

(b) 8832360163 

The reference angle for 163º, and thus for 883º, is 180º 163º = 17º.

Classroom Example 2 (page 58)

The reference angle for 225º is 225º 180º = 45º. 225º lies in quadrant III, so the tangent and cotangent are positive while the sine, cosine, secant, and cosecant are negative. 2



sin225sin45 2 2



cos225cos45 2

tan225tan451

cot225cot451

sec225sec452

csc225csc452

Classroom Example 3 (page 59)

(a) Because 150º is coterminal with an angle of 150º + 360º = 210º, it lies in quadrant III, and its sine is negative. The reference angle is 210º 180º = 30º. So  1 sin150sin30 2  .

(b) Because 780º is coterminal with an angle of 780236060  , it lies in quadrant I, and its cotangent is positive. The reference angle is 60º. So 3 cot780cot60 3   .

Classroom Example 4 (page 60) 22 sin45,cos135,tan2251 22 

Classroom Example 6 (page 61)

3 sin 2   sin  is negative, so  must lie in quadrants III or IV. The absolute value of sin  is 3 , 2 so the reference angle  must be 60º. The angle in quadrant III is 60180240,  and the angle in quadrant IV is 36060300.  

Section 2.3 Finding Trigonometric Function Values Using a Calculator

Classroom Example 1 (page 64)

(a) tan68432.56707352   (b) cos193.6220.97187064 

(d)   sec2873.42030362 

Classroom Example 2 (page 65)

(a) cos0.9211854122.899999  (b) cot1.446647434.654300 

Classroom Example 3 (page 66)

(a) sin5500sin3.9370 lb FW  

(b)   sin2800sin4.8230 lb FW   F is negative because the car is traveling downhill.

FW   



Classroom Example 5 (page 60)

Section 2.4 Solutions and Applications of Right Triangles

Classroom Example 1 (page 74)

902840906120ABAA  side adjacent25.3 cos hypotenuse 25.3

cos2840

25.3 28.8 cm (three cos2840 B AB AB cAB

significant digits) side opposite tan side adjacent25.3

tan2840 25.3 25.3tan284013.8 cm (three AC B AC bAC

  significant digits)

Classroom Example 2 (page 75)

Find the length of the hypotenuse, AB, using the Pythagorean theorem: 2222 55.8744.25 71.27 (four significant digits)

cABACBC   1 side opposite44.25 tan side adjacent55.87 44.25 tan38.38 or 3823 55.87 909038.3851.62 or 5137 A A BAB

Classroom Example 3 (page 76)

60.4 mB tantan60.4 15.515.5 15.5tan60.427.3 m bb

The tree is about 27.3 m tall.

Classroom Example 4 (page 76)

The angle of depression and angle B are alternate interior angles, so their measures are equal.

1 9797 tantan67 4242 BB 

Section 2.5 Further Applications of Right Triangles

Classroom Example 1 (page 83)

a a

cos538.6cos535.2 km 8.6

Classroom Example 2 (page 83)

The port is at A. Because the NS lines are parallel, the measure of angle ACD = 34°. Therefore, the measure of angle ACB = 34° + 56° = 90°, making triangle ABC a right triangle. The ship traveled at 18 knots for 2.5 hours or 18(2.5) = 45 nautical miles from A to C, labeled b in the figure. The ship traveled at 18 knots for 3.0 hours or 18(3.0) = 54 nautical miles from C to B, labeled a in the figure. Use the Pythagorean theorem to find the distance from B to A, labeled c in the figure. 222 2222 5445494170 abc ccc

The ship is about 70 nautical miles from port.

Classroom Example 3 (page 84)

(a) Find d when 24138 

and b = 3.5000 cm

From the figure, we see that 2 cotcot 222 b db d   . Convert  to decimal degrees: 241382.693888  

Then, cot 22 b d  

3.50002.693888 cot74.42708031 22 74.427 cm    d

(b) If  is 1 larger, then 241392.694167   

3.50002.694167 cot74.41940379 22   d

The difference is 74.4270803174.419403790.0076765 cm. 

Classroom Example 4 (page 84)

Algebraic solution:

There are two unknowns, the distance from the base of the building, x, and the height of the building, h.

In triangle ABC, tan74.2tan74.2. h hx x 

In triangle BCD, tan51.8 35 h x     35tan51.8. hx

Setting the two expressions equal, we have

tan74.235tan51.8 tan74.2tan51.835tan51.8 tan74.2tan51.835tan51.8 tan74.2tan51.835tan51.8 35tan51.8 tan74.2tan51.8 xx xx

tan74.2, hx so substitute the expression for x to find h: 35tan51.8 tan74.269 ft tan74.2tan51.8

Graphing calculator solution:

Superimpose coordinate axes on the figure with D at the origin. Thus, the coordinates of A are (35, 0). The tangent of the angle between the x-axis and the graph of a line with equation y = mx + b is the slope of the line. For line DB, m = tan 51.8º. Because b = 0, the equation of line DB is 1 tan51.8 yx . The equation of line AB is 2 tan74.2 yxb  . Use the coordinates of A and the point-slope form to find the equation of AB (continuedonnextpage)

(continued)

Graph 1y and 2y in the window [–10, 80] × [–10, 100], then find the point of intersection. The y-coordinate gives h.