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Chapter 2
Acute Angles and Right Triangles
Section 2.1 Trigonometric Functions of
Acute Angles
1.
21 20 21
; ;
29 29 20
2.
45 28 45
; ;
53 53 28
2 21 2 21 21 5 21 5
;
;
;
;
;
5 5
21
2
21 2
20. sin θ = cos (90-θ ); cos θ = sin (90-θ );
tan θ = cot (90-θ );cot θ = tan (90-θ );
sec θ = csc (90-θ ); csc θ = sec (90-θ )
21. sin 60
n m n
3.
; ;
p p m
4.
19. a = 21
22. cos 45
23. sec30
k y k
; ;
z z y
24. tan17
25. csc51
For Exercises 5−10, refer to the Function Values of
Special Angles chart on page 54 of the text.
26. cot 64.6
27. cos51.3
5. C
6. H
7. B
8. G
28. sin (70- )
9. E
10. A
29. csc (75- )
11. c = 13
12. c = 5
12 5 12 5 13 13
; ; ; ; ;
13 13 5 12 5 12
30. They are always the same values.
4 3 4 3 5 5
; ; ; ; ;
5 5 3 4 3 4
13. b = 13
13 6 13 6 13 7 7 13
; ;
;
; ;
7 7 6
13 6 17
7 95 7 95 95 12 95 12
14. a = 95
;
;
;
;
;
12 12
95
7
95
7
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. 40
32. 40
33. 20
34. 20
15. b = 91
91 3 91 3 91 10 10 91
; ;
;
; ;
10 10 3
91 3
91
35. 12
36. 12
37. 35
8 57 8 57 57 11 57 11
;
;
;
;
;
16. a = 57
11 11
57
8
57
8
38. 15
39. 18
17. b = 3
3 1
3
2 3
; ; 3;
; 2;
2 2
3
3
40. 35
18. b = 2
2 2
;
;1;1; 2; 2
2 2
42. true
41. true
43. false
Copyright © 2013 Pearson Education, Inc.
12
Section 2.1 Trigonometric Functions of Acute Angles
44. false
13
66.
45. true
46. false
47. true
48. false
49.
3
3
50.
3
51.
1
2
67. The legs of the right triangle provide the
(
coordinates of P, 2 2, 2 2
68.
52.
3
2
53.
2 3
3
54. 2
(1, 3)
69. sin x; tan x
70. cos x; csc x
71. 60
72. 0.7071067812 is a rational approximation for
the exact value
55.
2
56.
2
æ 2 2 ö÷
÷÷ ; 45°.
,
73. ççç
çè 2 2 ø÷
57.
2
2
74. y  3 x
58. 1
75. y =
59. 1
60.
61.
62.
2
(an irrational value).
2
3
x.
3
76. 30°
2
2
3
2
1
2
77. 60°
78. (a) 45º
(b)
2k
(c)
2
79. (a) 60
(b) k
63.
3
64.
2 3
3
65.
)
(c)
3k
(d) 2;
3; 30º; 60º
80. a = 12; b = 12 3; d = 12 3; c = 12 6
9
9 3
3 3
;z =
;w = 3 3
81. y = ; x =
2
2
2
82. m =
7 3
14 3
14 3
14 6
; a=
n=
; q=
.
3
3
3
3
83. p = 15; r = 15 2; q = 5 6; t = 10 6
Copyright © 2013 Pearson Education, Inc.
14
Chapter 2 Acute Angles and Right Triangles
84. A =
s2 3
4
85. A =
s2
2
7. 2 is a good choice for r because in a 30 - 60
right triangle, the hypotenuse is twice the
length of the shorter side (the side opposite to
the 30 angle). By choosing 2, one avoids
introducing a fraction (or decimal) when
determining the length of the shorter side.
Choosing any even positive integer for r
would have this result; however, 2 is the most
convenient value.
86. Yes
Section 2.2 Trigonometric Functions of
Non-Acute Angles
1. C
2. F
8. Answers may vary
3. A
9. Answers may vary
4. B
10. Answers may vary
5. D
6. B
θ
sin θ
cos θ
tan θ
3
3
1
11.
30
1
2
3
2
12.
45
2
2
13.
60
3
2
2
2
1
2
3
2
14. 120
2
2
15. 135
16. 150
sin150
= sin 30
1
=
2
3
cos 120
= - cos 60
=-
1
2
-
-
- 3
cot θ
sec θ
csc θ
3
2 3
3
2
1
2
3
3
cot 120
= - cot 60
3
3
tan135
cot135
= - tan 45 = - cot 45
= -1
= -1
=-
2
2
3
2
-
3
3
cot150
= - cot 30
=- 3
cos 210
17.
18.
210
240
-
-
1
2
3
2
= - cos 30
3
=2
-
1
2
3
3
tan 240
= tan 60
= 3
3
cot 240
= cot 60
3
=
3
2
2
2 3
3
sec120
= - sec 60
= -2
2 3
3
- 2
2
sec150
= - sec30
2
2 3
=3
sec 210
= - sec30
2 3
=3
-2
Copyright © 2013 Pearson Education, Inc.
-2
-
2 3
3
Section 2.2 Trigonometric Functions of Non-Acute Angles
19. -
3 1
3
2 3
; ; - 3; ; 2; 2 2
3
3
2 2
;
; -1; -1; 2; - 2
2 2
2 2
;
;1;1; 2; 2
2 2
3 1
3
2 3
; ; 3;
; 2;
2 2
3
3
3
2
38.
20. -
39. -
21.
40.
22.
3
42. - 2
3 1
3
2 3
; - ; - 3; ; -2; 2
2
3
3
24.
2
2
;; -1; -1; - 2; 2
2
2
43. - 1
3
3
44. -
1
3 3
2 3
;
; 3; ; -2
25. - ; 2
2 3
3
1 3 3
2 3
;
;
; 3;
;2
2 2 3
3
27. -
3
2
41. - 2
23.
26.
2
2
;;1;1; - 2; - 2
2
2
45. 1
46. 1
47.
23
4
48.
9
2
7
2
28.
3 1
3
2 3
; ; 3;
; 2;
2 2
3
3
49.
29.
3 1
3
2 3
; ; 3;
; 2;
2 2
3
3
50. 1
51. -
29
12
1 3
3
2 3
;; - 3;
; -2
30. - ;
2 2
3
3
52. 10
1
3 3
2 3
;
; 3; ; -2
31. - ; 2
2 3
3
53. Since 0 ¹
32.
33.
15
3 1
3
2 3
; ; 3;
; 2;
2 2
3
3
54. Since
1
3
3
2 3
;;; - 3; ;2
2
2
3
3
55. Since
3 +1
, the statement is false
2
1+ 3
¹ 1, the given statement is false
2
1
¹ 3, the statement is false
2
56. true
34. -
2
2
;;1;1; - 2; - 2
2
2
35. -
3 1
3
2 3
; ; - 3; ; 2; 2 2
3
3
36. -
2 2
;
; -1; -1; 2; - 2
2 2
37. -
2
2
57. true
58. true
59. false
60. true
61.
(-3
3,3 .
)
62.
(-5
2, -5 2 .
)
63. yes, in quadrant IV
Copyright © 2013 Pearson Education, Inc.
16
Chapter 2 Acute Angles and Right Triangles
64. there is no angle θ
89. 45;315
65. positive
90. 45;315
66. positive
Section 2.3 Finding Trigonometric
Function Values Using a
Calculator
67. positive
68. negative
69. negative
1. sin; 1
70. negative.
2. approximate
71. θ and θ + n ⋅ 360 are coterminal angles, so
the sine of each of these will result in the same
value.
3. reciprocal; reciprocal
72. θ and θ + n ⋅ 360 are coterminal angles, so
the cosine of each of these will result in the
same value.
73. If n is even, θ and θ + n ⋅180 are coterminal
angles, so the tangent of each of these will
result in the same value. If n is odd, θ and
θ + n ⋅180 have the same reference angle,
but are positioned two quadrants apart. The
tangent is positive for angles located in
quadrants I and III, and negative for angles
located in quadrants II and IV, so the tangent
of each angle is the same value.
4. sine; inverse
For Exercises 5−15, the calculation for decimal
degrees is indicated for calculators that do not accept
degree, minutes, and seconds.
5. sin 3842¢ » 0.62524266
6. cos 4124¢ » 0.75011107
7. sec1315¢ » 1.0273488
8. csc14545¢ » 1.7768146
9. cot18348¢ » 15.055723
10. tan 42130¢ » 1.8417709
11. sin (-31212 ¢) » 0.74080460
74. −0.4.
12. tan (-8006 ¢) » -5.7297416
75. sin 115° is closest to 0.9.
13. csc (-31736 ¢) » 1.4830142
76. 135; 315
77. 45; 225
14. cot (-51220 ¢) » 1.9074147
(b) 369 ft
15. tan 23.4 » 0.43273864
16. cos14.8 » 0.96682339
17. cot 77 » 0.23086819
(c) Answers will vary.
18. tan 33 » 0.64940759
78. (a) 550 ft
79. 30;150
80. 30;330
81. 60;300
82. 45; 225
83. 45;315
84. 60;300
85. 30;330
86. 60;300
87. 30;210 .
88. 60; 240
19. tan 4.72 » 0.08256640
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
sin 3.69 » 0.06435814
sin 39 » 0.629320391
tan 22 » 0.40402623









» 55.845496
» 81.168073
» 16.166641
» 57.997172
» 38.491580
» 46.173582
» 68.673241
» 30.502748
» 45.526434
Copyright © 2013 Pearson Education, Inc.
Section 2.3 Finding Trigonometric Functions Values Using a Calculator
32.
33.
34.
35.
 » 31.199998
 » 12.227282
 » 77.831359
A common mistake is to have the calculator in
radian mode, when it should be in degree
mode (and vice verse).
17
57. true
58. false
59. 65.96 lbs
60. -100.5lbs
61. » -2.9
62. » 2.9
63. » 2547 lbs
64. » 2771 lbs
65. The 2200-lb car on a 2° uphill grade has the
greater grade resistance.
66.
36. If the calculator allowed an angle θ where
0 £ θ < 360 , then we would need to find an
angle within this interval that is coterminal
with 2000º by subtracting a multiple of 360º:
2000 - 5 ⋅ 360 = 2000- 1800 = 200 . If
the calculator was more restrictive on
evaluating angles (such as 0 £ θ < 90 ), then
reference angles would need to be used.
37.
38.
39.
40.
41.
42.
43.
44.
45.
A = 56º.
A = 0.3746065934º
1
-1
1
0
0
1
A: 68.94 mph; B: 65.78 mph
46. r = a cos θ .
47. false
48. false
θ
sin θ
tan θ
0º
0.0000
0.0000
πθ
180
0.0000
0.5º
0.0087
0.0087
0.0087
1º
0.0175
0.0175
0.0175
1.5º
0.0262
0.0262
0.0262
2º
0.0349
0.0349
0.0349
2.5º
0.0436
0.0437
0.0436
3º
0.0523
0.0524
0.0524
3.5º
0.0610
0.0612
0.0611
4º
0.0698
0.0699
0.0698
(a) From the table, we see that if θ is small,
πθ
.
sin θ = tan θ =
180
(b) F = W sin θ = W tan θ =
W πθ
180
(c) 80 lb
(d) 117.81 lb
49. true
67. (a) 703 ft
50. true
(b) 1701 ft
(c) R will decrease; 644 ft; 1559 ft.
68. 55 mph
69. (a) 2 ´108 m per sec.
51. false
52. false
53. false
54. false
55. true
56. true
(b) 2 ´108 m per sec.
70. (a) 19
(b) 50
71. 48.7
Copyright © 2013 Pearson Education, Inc.
18
Chapter 2 Acute Angles and Right Triangles
4. Answers will vary.
No; the number of points scored will be a
whole number.
72. 40.8; 7.9
73. (a) 155 ft
(b) 194 ft
5. Answers will vary.
It would be cumbersome to write 2 as 2.00 or
2.000, for example, if the measurements had 3
or 4 significant digits (depending on the
problem). In the formula, it is understood that
2 is an exact value. Since the radius
measurement, 54.98 cm, has four significant
digits, an appropriate answer would be 345.4
cm.
74. 78 mph
Chapter 2 Quiz
(Sections 2.1−2.3)
3 4 3 4 5 5
; ; ; ; ;
5 5 4 3 4 3
1.
2. θ
sin θ
cos θ
tan θ
30
1
2
3
2
3
3
45
2
2
60
3
2
2
2
1
2
cot θ
sec θ
csc θ
3
2 3
3
2
1
1
2
3
3
3
2
2
2 3
3
6. 23.0 ft indicates 3 significant digits and 23.00
ft indicates four significant digits.
7. 0.05
8. 0.5
9. B = 5340 ¢; a » 571m; b » 777 m
10. Y = 42.2; x » 66.4 cm; y » 60.2 cm
11. M = 38.8; n » 154 m; p » 198 m
3. w = 18; x = 18 3; y = 18; z = 18 2
12. B = 5820¢; c » 68.4 km; b » 58.2 km
4. 3x2 sin  .
13.
5.
A = 47.9108; c » 84.816 cm;
a » 62.942 cm
14. A = 21.4858; b » 3330.68 m;
a » 1311.04 m
2
2
;; -1; -1; - 2; 2
2
2
1
3 3
2 3
;
; 3; ; -2
6. - ; 2
2 3
3
15. c » 20.5 ft; A » 3740 ¢; B » 5220¢
3 1
3
2 3
; ; - 3; ; 2; 2 2
3
3
8. 60;120 .
17. No; You need to have at least one side to solve
the triangle.
9. 45; 225
19. 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.
7. -
10.
11.
12.
13.
14.
0.67301251
-1.1817633
 » 69.497888
 » 24.777233
false
15. true
Section 2.4
Solving Right Triangles
1. 22,894.5 to 22,895.5
2. 28,999.5 to 29,000.5
3. 8958.5 to 8959.5
16. b » 14.5 m; A » 1820 ¢; B » 7140¢
18. The other acute angle requires the least work.
20. Answers will vary. If you know the lengths of
two sides, you can set up a trigonometric ratio
to solve for one of the acute angles. The other
acute angle may be found by subtracting the
calculated acute angle from 90°. With either of
the two acute angles that have been
determined, you can set up a trigonometric
ratio along with one of the known sides to
solve for the missing side.
21. B = 62.0; a » 8.17 ft; b » 15.4 ft
22. A = 44.0; a » 20.6 m; b » 21.4 m
Copyright © 2013 Pearson Education, Inc.
Section 2.5 Further Applications of Right Triangles
23. A = 17.0; a = 39.1 in;c = 134 in
24. B = 27.5; b » 6.61 m;c » 14.3 m
25. B = 29.0; b » 70.7 cm;c » 80.9 cm
26.
A = 38.3; b » 35.6 ft;c » 45.3 ft
27. b » 18 m;
A » 36; B » 54
29. c » 85.9 yd; A » 6250¢; B » 2710¢
30. c » 1080 m; A » 6300¢; B » 2700¢
31. b » 42.3 cm; A » 2410 ¢; B » 6550 ¢
32. a » 609 cm; A » 7010¢; B » 1950¢
33. B » 3636 ¢; b » 310.8 ft;c » 230.8 ft
34. B » 7613¢; a » 306.2 m;b » 1248 m
35. B » 5051¢; b » 0.3934 m;a » 0.4832 m
36. B = 709 ¢; b » 4.787 m; a » 0.6006 m
37. The angle of elevation from X to Y (with Y
above X) is the acute angle formed by ray XY
and a horizontal ray with endpoint at X.
38. No
39. Answers will vary. The angle of elevation and
the angle of depression are measured between
the line of sight and a horizontal line. So, in
the diagram, lines AD and CB are both
horizontal. Hence, they are parallel. The line
formed by AB is a transversal and angles DAB
and ABC are alternate interior angle and thus
have the same measure.
40. The angle of depression is measured between
the line of sight and a horizontal line. This
angle is measured between the line of sight
and a vertical line.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
33.4 m
128 ft
864,900 mi
26.92 in
134.7 cm
22
583 ft.
28.0 m
469 m
13.3 ft
42,600 ft
3735¢
146 m
42.18
63.39
(a) 29,000 ft.
(b) shorter
58. 34.0 mi
28. a » 40 ft; A » 51; B » 39
41. 9.35 m
53.
54.
55.
56.
57.
19
Section 2.5 Further Applications of Right
Triangles
1. It should be shown as an angle measured
clockwise from due north.
2. It should be shown measured from north (or
south) in the east (or west) direction.
3. A sketch is important to show the relationships
among the given data and the unknowns.
4. The angle of elevation (or depression) from X
to Y is measured from the horizontal line
through X to the ray XY.
5. 270°; N 90° W, or S 90° W
6.
7.
8.
9.
10.
11.
12.
90°; N 90° E, or S 90° E
0°; N 0° E or N 0º W
180°; S 0° E or S 0º W
315°; N 45° W
225°; S 45° W
135°; S 45° E
45°; N 45° E
3
3
13.
y=
14.
y = - 3 x, x ³ 0
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
x, x £ 0
220 mi
150 km
47 nautical mi
70 nautical mi
2203 ft
5856 m
148 mi
2.01 mi
430 mi
350 mi
140 mi
130 mi
27. x =
b
a -c
Copyright © 2013 Pearson Education, Inc.
20
Chapter 2 Acute Angles and Right Triangles
28. tan θ = - tan (180-θ ) = - tan θ ¢ . This is
because the angle represented by 180- θ
terminates in quadrant II if 0 < θ < 90 . If
90 < θ < 180 , then the angle represented by
180- θ terminates in quadrant I. Thus,
tan θ and tan (180-θ ) are opposite in sign.
The slope of the line is m = - ba , and
tan θ = - tan (180-θ ) = - tan θ ¢ = - ba .
Thus, m = - ba = tan θ . The point-slope form
of the equation of a line is y - y1 = m ( x - x1 )
Substituting tan θ for m into
y - y1 = m ( x - x1 ) , we have
44. (a)As θ increases, D increases and then
decreases.
(b) 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.
Chapter 2
Review Exercises
60 11 60 11 61 61
; ; ; ; ;
61 61 11 60 11 60
20 21 20 21 29 29
2.
; ; ; ; ;
29 29 21 20 21 20
1.
3. 10
y - y1 = - tan θ ( x - x1 ) .
The line passes through (a, 0), so
y - y1 = tan θ ( x - x1 ) 
y - 0 = tan θ ( x - a )  y = tan θ ( x - a) .
4. 10
5. 7
6. 30
7. true
29. y = tan 35 ( x - 25)
8. false
30. y = tan15 ( x - 5)
9. true
31. 433 ft
32. 448 m
33. 114 ft
10. false
11. The sum of the measures of angles A and B is
90, and, thus, they are complementary
angles. Since sine and cosine are cofunctions,
we have sin B = cos (90- B) = cos A.
34. 147 m
35. 5.18 m
36. 2.47 km
12. D, tan 140°, is the only one which cannot be
determined exactly using the methods of this
chapter.
3 1
3
2 3
13. - ; ; - 3; - ; 2; 2 2
3
3
bæ 
ö
37. (a) d = ççcot + cot ÷÷÷
2 çè
2
2ø
(b) 345.4 cm
38. 1.95 mi
39. 10.8 ft
14.
40. A » 35.987 » 3559 ¢10 ¢¢
B » 54.013 » 5400 ¢50 ¢¢
3 1
3
2 3
; - ; - 3; ; -2;
2
2
3
3
1 3
3
2 3
;; - 3;
; -2
15. - ;
2 2
3
3
41. (a) 320 ft
æ
ö
(b) R çç1- cos ÷÷÷
çè
2ø
16.
42. 84.7 m
43. (a) 23 ft
18. 30;330
2
2
;; -1; -1; - 2; 2
2
2
17. 60; 240
19. 30;210
(b) 48 ft
(c) The faster the speed, the more land needs
to be cleared on the inside of the curve.
20. 45;315
21. 3 -
2 3
3
Copyright © 2013 Pearson Education, Inc.
Chapter 2 Test
22. 1
23.
60. 110 km
61. 140 mi
7
2
62. k ( tan B - tan A)
63.−64. Answers will vary.
2
2
24. (a) ;;1
2
2
65. (a) 716 mi
(b) 1104 mi
3 1
; ;- 3
2 2
-1.3563417
0.95371695
1.0210339
-0.71592968
0.20834446
1.9362132
55.673870°
41.635092°
12.733938°
37.695528°
63.008286°
5.9998273°
47.1; 132.9
54.2º; 234.2º
false
true
(b) -
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
66. (a) xQ = xP + d sin  ; yQ = yP + d cos  .
(b) (181.34, 523.02)
Chapter 2
1.
4. (a) true
(b) false
(c) true
5. -
2
2
;;1;1; - 2; - 2
2
2
7. -1;0; undefined;0; undefined; -1
8. 135; 225
11. Take the reciprocal of tan  to get
cot   0.59600119.
48. quadrant I
49. B = 3130 ¢; a » 638; b » 391
50. c » 390.3; A » 1925¢; B » 7035¢
51. B = 50.28; a » 32.38 m; c » 50.66 m
52. A = 4207 ¢; a » 270.0 m; c » 402.5 m
55.
56.
57.
58.
59.
73.7 ft
20.4 m
18.75 cm
50.24 m
1200 m
9. 240; 300
10. 45; 225
Answers may vary.
quadrant III
quadrant II
quadrant II
54. r = 13; » 23
3 1
3
2 3
; - ; 3;
; -2; 2
2
3
3
6. -
1
¹ tan-1 25.
tan 25
53. 137 ft
12 5 12 5 13 13
; ; ; ; ;
13 13 5 12 5 12
3. 15
42. false
44.
45.
46.
47.
Chapter Test
2. y = 4 3; w = 8; x = 4; z = 4 2
41. true
43. No, cot 25 =
21
12. (a) 0.97939940
(b) -1.9056082
(c) 1.9362432
13. θ » 16.166641
14. B = 3130¢; b » 458; c » 877
15.
16.
17.
18.
19.
20.
67.1 or 6710¢
15.5 ft
8800 ft
72 nautical mi
92 km
448 m
Copyright © 2013 Pearson Education, Inc.
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