Download Review Exercises for Chapter 6

592
Chapter 6
Additional Topics in Trigonometry
Review Exercises for Chapter 6
1. Given: A 35, B 71, a 8
C 180 35 71 74
2. Given: A 22, B 121, a 17
C 180 A B 37
b
a sin B 8 sin 71
13.19
sin A
sin 35
b
a sin B 17 sin 121
38.90
sin A
sin 22
c
a sin C 8 sin 74
13.41
sin A
sin 35
c
a sin C 17 sin 37
27.31
sin A
sin 22
3. Given: B 72, C 82, b 54
A 180 72 82 26
4. Given: B 10, C 20, c 33
A 180 B C 150
a
b sin A 54 sin 26
24.89
sin B
sin 72
a
c sin A 33 sin 150
48.24
sin C
sin 20
c
b sin C 54 sin 82
56.23
sin B
sin 72
b
c sin B 33 sin 10
16.75
sin C
sin 20
5. Given: A 16, B 98, c 8.4
C 180 16 98 66
6. Given: A 95, B 45, c 104.8
C 180 A B 40
a
c sin A 8.4 sin 16
2.53
sin C
sin 66
a
c sin A 104.8 sin 95
162.42
sin C
sin 40
b
c sin B 8.4 sin 98
9.11
sin C
sin 66
b
c sin B 104.8 sin 45
115.29
sin C
sin 40
7. Given: A 24, C 48, b 27.5
B 180 24 48 108
8. Given: B 64, C 36, a 367
A 180 B C 80
a
b sin A 27.5 sin 24
11.76
sin B
sin 108
b
a sin B 367 sin 64
334.95
sin A
sin 80
c
b sin C 27.5 sin 48
21.49
sin B
sin 108
c
a sin C 367 sin 36
219.04
sin A
sin 80
9. Given: B 150, b 30, c 10
sin C c sin B 10 sin 150
0.1667 ⇒ C 9.59
b
30
A 180 150 9.59 20.41
b sin A 30 sin 20.41
a
20.92
sin B
sin 150
11. A 75, a 51.2, b 33.7
sin B b sin A 33.7 sin 75
0.6358 ⇒ B 39.48
a
51.2
C 180 75 39.48 65.52
c
a sin C 51.2 sin 65.52
48.24
sin A
sin 75
10. Given: B 150, a 10, b 3
sin A a sin B 10 sin 150
1.67 > 1
b
3
No solution
Review Exercises for Chapter 6
593
12. Given: B 25, a 6.2, b 4
sin A a sin B
0.65506 ⇒ A 40.92 or 139.08
b
Case 1: A 40.92
Case 2: A 139.08
C 180 25 40.92 114.08
C 180 25 139.08 15.92
c 8.64
c 2.60
13. Area 12bc sin A 1257sin 27 7.9
14. B 80º, a 4, c 8
Area 12ac sin B 12480.9848 15.8
1
1
15. Area 2ab sin C 2165sin 123 33.5
16. A 11, b 22, c 21
Area 12 bc sin A 12 22210.1908 44.1
17. tan 17 h
⇒ h x 50 tan 17
x 50
h
h x tan 17 50 tan 17
tan 31 31°
x
17°
50
h
⇒ h x tan 31
x
x tan 17 50 tan 17 x tan 31
50 tan 17 xtan 31 tan 17
50 tan 17
x
tan 31 tan 17
x 51.7959
h x tan 31 51.7959 tan 31 31.1 meters
The height of the building is approximately 31.1 meters.
18. 162 w2 122 2w12 cos 140
w2 24 cos 140w 112 0 ⇒ w 4.83
19.
h
75
sin 17 sin 45
75 sin 17
h
sin 45
45°
118°
h 31.01 feet
ft
75
62°
17°
28°
20. The triangle of base 400 feet formed by the two angles of sight to the tree has base angles
of 90 22 30 67 30, or 67.5, and 90 15 75. The angle at the tree measures
180 67.5 75 37.5.
400 sin 75
b
634.683
sin 37.5
h 634.683 sin 67.5
h 586.4
The width of the river is about 586.4 feet.
45°
Tree
A
N
W
E
S
15°
h
22° 30'
C
400 ft
B
h
594
Chapter 6
Additional Topics in Trigonometry
21. Given: a 5, b 8, c 10
a2
cos C 2ab
b2
c2
22. Given: a 80, b 60, c 100
0.1375 ⇒ C 97.90
a2 c2 b2
0.61 ⇒ B 52.41
2ac
cos B a2 b2 c2 6400 3600 10,000
2ab
28060
cos C 0 ⇒ C 90
A 180 B C 29.69
sin A 80
0.8 ⇒ A 53.13º
100
sin B 60
0.6 ⇒ B 36.87º
100
23. Given: a 2.5, b 5.0, c 4.5
cos B a2 c2 b2
0.0667 ⇒ B 86.18
2ac
cos C a2 b2 c2
0.44 ⇒ C 63.90
2ab
A 180 B C 29.92
24. Given: a 16.4, b 8.8, c 12.2
cos A b2 c2 a2 8.82 12.22 16.42
0.1988 ⇒ A 101.47
2bc
28.812.2
sin B b sin A 8.8 sin 101.47
0.5259 ⇒ B 31.73
a
16.4
C 180 101.47 31.73 46.80
25. Given: B 110, a 4, c 4
b
a2
c2
26. Given: B 150, a 10, c 20
2ac cos B 6.55
A C 12 180 110 35
b2 102 202 21020cos 150 ⇒ b 29.09
sin A a sin B 10 sin 150
⇒ A 9.90
b
29.09
C 180 150 9.90 20.10
27. Given: C 43, a 22.5, b 31.4
c a2 b2 2ab cos C 21.42
cos B a2 c2 b2
0.02169 ⇒ B 91.24
2ac
A 180 B C 45.76
28. Given: A 62, b 11.34, c 19.52
a2 11.342 19.522 211.3419.52 cos 62 ⇒ a 17.37
sin B b sin A 11.34 sin 62
⇒ B 35.20
a
17.37
C 180 62 35.20 82.80
Review Exercises for Chapter 6
29.
5 ft
8 ft
8 ft
28°
5 ft
152°
30.
595
15 m
a
20 m
b
20 m
34°
15 m
146°
s1
s2
a2 52 82 258cos 28 18.364
a 4.3 feet
b2 82 52 285cos 152 159.636
b 12.6 feet
s12 152 202 2 15
20 cos 34 127.58
s1 11.3 meters
s22 15 2 202 2 15
20 cos 146 1122.42
s2 33.5 meters
31. Length of AC 3002 4252 2300425 cos 115
32. d 2 8502 10602 28501060 cos 72
1,289,251
615.1 meters
d 1135 miles
N
W
d
5°
E
S
850
67°
33. a 4, b 5, c 7
s
1060
34. a 15, b 8, c 10
abc 457
8
2
2
s
15 8 10
16.5
2
Area 16.51.58.56.5 36.979
Area ss as bs c
8431 9.80
35. a 12.3, b 15.8, c 3.7
s
36. a 38.1, b 26.7, c 19.4
a b c 12.3 15.8 3.7
15.9
2
2
Area ss as bs c
s
38.1 26.7 19.4
42.1
2
Area 42.1415.422.7 242.630
15.93.60.112.2 8.36
37. u 4 22 6 12 61
v 6 02 3 22 61
38. u 3 12 2 42 210
v 1 32 4 22 210
u is directed along a line with a slope of
61
5
.
4 2 6
u is directed along a line with a slope of
2 4
3.
31
v is directed along a line with a slope of
3 2 5
.
60
6
v is directed along a line with a slope of
4 2
3.
1 3
Since u and v have identical magnitudes and directions,
u v.
39. Initial point: 5, 4
Since u and v have identical magnitudes and directions,
u v.
40. Initial point: 0, 1
Terminal point: 2, 1
7
Terminal point: 6, 2 v 2 5, 1 4 7, 5
v 6 0, 72 1 6, 52
596
Chapter 6
Additional Topics in Trigonometry
41. Initial point: 0, 10
42. Initial point: 1, 5
Terminal point: 7, 3
Terminal point: 15, 9
v 7 0, 3 10 7, 7
v 15 1, 9 5 14, 4
43. v 8, 120
8 cos 120, 8 sin 120 4, 43
45. u 1, 3, v 3, 6
1
44. v , 225
2
12 cos 225, 21 sin 225 42, 42
46. u 4, 5, v 0, 1
(a) u v 1, 3 3, 6 4, 3
(a) u v 4 0, 5 1 4, 4
(b) u v 1, 3 3, 6 2, 9
(b) u v 4 0, 5 1 4, 6
(c) 3u 31, 3 3, 9
(c) 3u 34, 35 12, 15
(d) 2v 5u 23, 6 51, 3
(d) 2v 5u 20, 21 54, 55
6, 12 5, 15 11, 3
47. u 5, 2, v 4, 4
0 20, 2 25 20, 23
48. u 1, 8, v 3, 2
(a) u v 5, 2 4, 4 1, 6
(a) u v 1 3, 8 2 4, 10
(b) u v 5, 2 4, 4 9, 2
(b) u v 1 3, 8 2 2, 6
(c) 3u 35, 2 15, 6
(c) 3u 31, 38 3, 24
(d) 2v 5u 24, 4 55, 2
(d) 2v 5u 23, 22 51, 58
8, 8 25, 10 17,18
49. u 2i j, v 5i 3j
6 5, 4 40 11, 44
50. u 7i 3j, v 4i j
(a) u v 2i j 5i 3j 7i 2j
(a) u v 7i 3j 4i j 3i 4j
(b) u v 2i j 5i 3j 3i 4j
(b) u v 7i 3j 4i j 11i 2j
(c) 3u 32i j 6i 3j
(c) 3u 37i 3j 21i 9j
(d) 2v 5u 25i 3j 52i j
(d) 2v 5u 8i 2j 35i 15j
10i 6j 10i 5j 20i j
51. u 4i, v i 6j
27i 17j
52. u 6j, v i j
(a) u v 4i i 6j 3i 6j
(a) u v 6j i j i 5j
(b) u v 4i i 6j 5i 6j
(b) u v 6j i j i 7j
(c) 3u 34i 12i
(c) 3u 18j
(d) 2v 5u 2i 6j 54i
(d) 2v 5u 2i 2j 30j
2i 12j 20i 18i 12j
2i 28j
Review Exercises for Chapter 6
53. u 6i 5j, v 10i 3j
597
54. u 6i 5j, v 10i 3j
2u v 26i 5j 10i 3j
22i 7j
4u 5v 24i 20j 50i 15j
26i 35j
y
v
2
22, 7
26,35
x
−5
10
−2
y
20
20 25 30
x
− 60
−4
− 40
20
−5v
4u
−6
−8
2u + v
2u
4u − 5v − 40
− 10
− 60
− 12
55.
v 10i 3j
56. v 10i 3j
y
3v 310i 3j
1
2v
20
30i 9j
y
5i 32j 5, 32
8
6
10
30, 9
3v
4
v
v
2
x
10
20
30
1
v
2
x
2
− 10
57. u 3, 4 3i 4j
58. u 6, 8 6i 8j
59. Initial point: 3, 4
60. Initial point: 2, 7
6
Terminal point: 5, 9
u 9 3i 8 4j 6i 4j
u 5 2, 9 7 7, 16 7i 16j
62. v 4i j
v 10 10 200 102
2
tan 2
10
1 ⇒ 135 since
10
v is in Quadrant II.
v 102i cos 135 j sin 135
v 7cos 60 i sin 60 j
v 42 12 17
tan 1
, in Quadrant IV ⇒ 346
4
v 17cos 346 i sin 346 j
64. v 3cos 150i sin 150 j
v 3, 150
v 7
60
65.
66. v 4i 7j
v 5i 4j
v 52 42 41
tan 67.
8
Terminal point: 9, 8
61. v 10i 10j
63.
4
−2
4
⇒ 38.7
5
v 3i 3j
tan tan 7
, in Quadrant II ⇒ 119.7
4
68. v 8i j
v 3 3 32
2
v 42 72 65
2
3
1 ⇒ 225
3
v 82 12 65
tan 1
, in Quadrant IV ⇒ 352.9
8
10
598
Chapter 6
Additional Topics in Trigonometry
69. Magnitude of resultant:
70. Rope One:
23i 21j
c 852 502 28550 cos 165
u ucos 30i sin 30j u
133.92 pounds
Rope Two:
Let be the angle between the resultant and the 85-pound
force.
cos v u cos 30i sin 30j u 133.92 85 50
2133.9285
2
2
2
2
1
i j
2
Resultant: u v uj 180j
0.9953
u 180
⇒ 5.6
Therefore, the tension on each rope is u 180 lb.
71. Airspeed: u 430cos 45i sin 45j 2152i j
Wind: w 35cos 60 sin 60 j y
u w N
135° W
35
i 3j
2
θ
35
353
i
2152
Groundspeed: u w 2152 2
2
215 2 35
2
2
E
S
x
45°
u
353
2152
2
2
w
422.30 miles per hour
Bearing: tan 17.53 2152
2152 17.5
40.4
90 130.4
72. Airspeed: u 724cos 60i sin 60j
y
362i 3j
Wind: w 32i
Groundspeed u w 394i 3623j
u w 3942 36232 740.5 kmhr
tan 3
3623
⇒ 57.9
394
724
30°
x
32
Bearing: N 32.1 E
73. u 6, 7, v 3, 9
u v 63 79 45
75. u 3i 7j, v 11i 5j
u v 311 75 2
77. u 3, 4
2u 6, 8
2u u 63 84 50
The result is a scalar.
74. u 7, 12, v 4, 14
u v 74 1214 140
76. u 7i 2j, v 16i 12j
u v 716 212 136
78. v 2, 1
v2 v v 22 12 5; scalar
Review Exercises for Chapter 6
79. u 3, 4, v 2, 1
599
80. u 3, 4, v 2, 1
u v 32 41 2
3u v 332 41 32 6; scalar
uu v u2 2u 6, 8
The result is a vector.
81. u cos
7
7
1
1
,
i sin j 2
2
4
4
82. u cos 45 i sin 45 j
v cos 300 i sin 300 j
v cos
3 1
5
5
i sin j ,
6
6
2 2
cos 3 1
uv
11
⇒ u v
22
12
Angle between u and v: 60 45 105
84. u 3, 3 , v 4, 33 83. u 22, 4, v 2, 1
cos uv
8
⇒ 160.5
u v 243
86. u 85. u 3, 8
cos uv
21
⇒ 22.4
u v 1243
14, 12, v 2, 4
87. u i
v 8u ⇒ Parallel
v 8, 3
v i 2j
u v 38 83 0
u
u and v are orthogonal.
v ku ⇒ Not parallel
v0
⇒ Not orthogonal
Neither
88. u 2i j, v 3i 6j
u
v0
⇒ Orthogonal
89. u 4, 3, v 8, 2
w1 projvu v v 688, 2 174, 1
u
v
26
w2 u w1 4, 3 u w1 w2 90. u 5, 6, v 10, 0
w1 projvu 50
10, 0 5, 0
uv vv 100
2
13
2
13
16
4, 1 1, 4
17
17
13
16
4, 1 1, 4
17
17
91. u 2, 7, v 1, 1
w1 projvu v v 2 1, 1
u
v
5
2
w2 u w1 5, 6 5, 0 0, 6
5
1, 1
2
u w1 w2 5, 0 0, 6
w2 u w1 2, 7 521, 1
9
1, 1
2
5
9
u w1 w2 1, 1 1, 1
2
2
600
Chapter 6
Additional Topics in Trigonometry
\
93. P 5, 3, Q 8, 9 ⇒ PQ 3, 6
92. u 3, 5, v 5, 2
w1 projvu uv
25
v 5, 2
v2
29
\
w2 u w1 3, 5 u w1 w2 W v PQ 2, 7
3, 6 48
25
19
5, 2 2, 5
29
29
19
25
5, 2 2, 5
29
25
95. w 18,00048
12 72,000 foot-pounds
94. work v PQ
\
3i 6j 10i 17j
30 102
132
97. 7i 02 72 7
\
96. W cos F PQ Imaginary
axis
cos 2025 pounds12 ft
10
281.9 foot-pounds
8
7i
6
4
2
−6
98. 6i 6
99. 5 3i 52 32
Imaginary
axis
34
8
−2
2
4
Imaginary
axis
5 + 3i
3
2
4
6
8
Real
axis
2
1
−4
−6
−6i
−1
−1
−8
100. 10 4i 102 42
229
tan 4
2
102.
z 5 12i
z tan 52
2
3
r 52 52 50 52
6
− 10 − 4i
1
101. 5 5i
Imaginary
axis
−12 − 10 −8 − 6
Real
axis
5
7
1 ⇒ since the
5
4
complex number is in Quadrant IV.
−2
−4
5 5i 52 cos
−6
7
7
i sin
4
4
103. 33 3i
122
13
12
⇒ 1.176
5
z 13cos 1.176 i sin 1.176
Real
axis
−2
4
4
2
6
5
6
−8 −6 −4 −2
−4
r 33 2 32 36 6
tan 3
1
5
⇒ 3
33
6
since the complex number is in Quadrant II.
33 3i 6 cos
5
5
i sin
6
6
4
5
Real
axis
Review Exercises for Chapter 6
601
z 7
104.
z 7
tan 0
0 ⇒ 7
z 7cos i sin 105. (a) z1 23 2i 4 cos
(a ) z2 10i 10 cos
3
3
i sin
2
2
11
11
i sin
6
6
10
10
i sin
3
3
(b) z1z2 4 cos
40 cos
z1
z2
11
11
i sin
6
6
z2 23 i 4 cos
5
5
i sin
4
4
i sin
6
6
5
5
i sin
4
4
17
17
i sin
12
12
122 cos
5
(b) z1z2 32 cos
3
3
i sin
2
2
10
cos
107. 5 cos
11
11
i sin
6
6
2
cos i sin
3
3
5
3
3
i sin
10 cos
2
2
4 cos
106. (a) z1 31 i 32 cos
z1
z2
4
cos 6 i sin 6 5
4 i sin 4 32 13
13
cos
i sin
4 12
12 4 cos i sin 6
6
32 cos
i sin
12
12
4
54 cos
4
4
i sin
12
12
625 cos
625
i sin
3
3
108.
4
4
2
cos 15 i sin 15 5
25 cos
32 12 23i
4
4
i sin
3
3
1 3
i
2
2
16 163i
625 6253
i
2
2
109. 2 3i6 13cos 56.3 i sin 56.36
110. 1 i8 2cos 315 i sin 315
8
133cos 337.9 i sin 337.9
16cos 2520 i sin 2520
13 0.9263 0.3769i
16cos 0 i sin 0
2035 828i
16
3
602
Chapter 6
Additional Topics in Trigonometry
111. Sixth roots of 729i 729 cos
3
3
:
i sin
2
2
(a) and (c)
(b)
6
729 cos
3
2k
2
6
i sin
3
2k
2
6
4
, k 0, 1, 2, 3, 4, 5
32 32
k 0: 3 cos i sin
i
4
4
2
2
7
7
0.776 2.898i
i sin
12
12
11
11
i sin
2.898 0.776i
12
12
5
5
32 32
i sin
i
4
4
2
2
19
19
i sin
0.776 2.898i
12
12
23
23
i sin
2.898 0.776i
12
12
k 1: 3 cos
k 2: 3 cos
k 3: 3 cos
k 4: 3 cos
k 5: 3 cos
Imaginary
axis
−4
−2
4
−2
−4
112. (a) 256i 256 cos
i sin
2
2
(b)
Imaginary
axis
5
Fourth roots of 256i:
3
2k
2k
2
2
4 256 cos
i sin
, k 0, 1, 2, 3
4
4
i sin
8
8
5
5
i sin
8
8
9
9
i sin
8
8
13
13
i sin
8
8
k 0: 4 cos
k 1: 4 cos
k 2: 4 cos
k 3: 4 cos
1
−3
−1
1 2 3
−2
−3
−5
(c) 3.696 1.531i
1.531 3.696i
3.696 1.531i
1.531 3.696i
113. Cube roots of 8 8cos 0 i sin 0, k 0, 1, 2
(a) and (c)
3
8 cos
(b)
0 2k
0 2k
i sin
3
3
Imaginary
axis
3
k 0: 2cos 0 i sin 0 2
2
2
k 1: 2 cos
i sin
1 3i
3
3
4
4
i sin
1 3i
3
3
k 2: 2 cos
Real
axis
−3
−1
1
−3
3
Real
axis
5
Real
axis
Review Exercises for Chapter 6
114. (a) 1024 1024cos i sin (b)
Imaginary
axis
Fifth roots of 1024:
5 1024 cos
5
2k
2k
, k 0, 1, 2, 3, 4
i sin
5
5
k 0: 4 cos i sin
5
5
k 1: 4 cos
3
3
i sin
5
5
7
7
i sin
5
5
9
9
i sin
5
5
k 4: 4 cos
2 3
Real
axis
5
−5
(c) 3.236 2.351i
k 2: 4cos i sin k 3: 4 cos
1
−3 −2 −1
1.236 3.804i
4
1.236 3.804i
3.236 2.351i
115. x4 81 0
x4 81
Solve by finding the fourth roots of 81.
81 81cos i sin 4 81 4 81 cos
2k
2k
i sin
, k 0, 1, 2, 3
4
4
Imaginary
axis
4
32 32
k 0: 3 cos i sin
i
4
4
2
2
3
3
32 32
i sin
i
4
4
2
2
5
5
32 32
i sin
i
4
4
2
2
7
7
32 32
i sin
i
4
4
2
2
k 1: 3 cos
k 2: 3 cos
k 3: 3 cos
2
−4
−2
2
Real
axis
4
−2
−4
116. x5 32 0
Imaginary
axis
x5 32
3
32 32cos 0 i sin 0
3 32 5 32 cos 0 2k
2k
i sin 0 5
5
1
−3
−1
k 0, 1, 2, 3, 4
k 0: 2cos 0 i sin 0 2
2
2
i sin
0.6180 1.9021i
5
5
4
4
i sin
1.6180 1.1756i
5
5
6
6
i sin
1.6180 1.1756i
5
5
8
8
i sin
0.6180 1.9021i
5
5
k 1: 2 cos
k 2: 2 cos
k 3: 2 cos
k 4: 2 cos
−3
1
3
Real
axis
603
604
Chapter 6
Additional Topics in Trigonometry
117. x3 8i 0
x3
Imaginary
axis
8i
Solve by finding the cube roots of 8i.
3
3
8i 8 cos
i sin
2
2
3 8i 3 8 cos
3
2k
2
3
1
i sin
i sin
2i
2
2
7
7
i sin
3 i
6
6
11
11
i sin
3 i
6
6
k 0: 2 cos
k 1: 2 cos
k 2: 2 cos
3
3
2 k
2
3
−3
, k 0, 1, 2
3
−1
Real
axis
−3
118. x3 1x2 1 0
Imaginary
axis
x3 1 0
2
x2 1 0
x3 1
−2
2
Real
axis
1 1cos 0 i sin 0
3
3
1 1 cos
0 2k
0 2k
i sin
3
3
, k 0, 1, 2
−2
1cos 0 i sin 0 1
2
2
1 3
i sin
i
3
3
2
2
4
4
1 3
i sin
i
3
3
2
2
1 cos
1 cos
x2 1 0
x2 1
1 1cos i sin 1 1 cos
2k
2k
i sin
, k 0, 1
2
2
k 0, 1
i sin
i
2
2
3
3
i sin
i
2
2
1 cos
1 cos
119. True. sin 90 is defined in the Law of Sines.
120. False. There may be no solution, one solution, or
two solutions.
121. True, by the definition of a unit vector.
v
so v vu
u
v
122. False, a b 0.
Review Exercises for Chapter 6
123. False. x 3 i is a solution to x3 8i 0, not
x2 8i 0.
124.
a
b
c
sin A sin B sin C
or
605
sin A sin B sin C
a
b
c
Also, 3 i2 8i 2 23 8i 0.
125. a2 b2 c2 2bc cos A
b2 a2 c2 2ac cos B
126. A vector in the plane has both a magnitude and
a direction.
c2 a2 b2 2ab cos C
127. A and C appear to have the same magnitude and direction.
128. u v is larger in figure (a) since the angle between
u and v is acute rather than obtuse.
129. If k > 0, the direction of ku is the same, and the
magnitude is ku.
130. The sum of u and v lies on the diagonal of the
parallelogram with u and v as its adjacent sides.
If k < 0, the direction of ku is the opposite direction of
u, and the magnitude is k u.
131. (a) The trigonometric form of the three roots shown is:
4cos 60 i sin 60
132. (a) The trigonometric forms of the four roots shown are:
4cos 60 i sin 60
4cos 180 i sin 180
4cos 150 i sin 150
4cos 300 i sin 300
(b) Since there are three evenly spaced roots on the circle of radius 4, they are cube roots of a complex
number of modulus 43 64.
Cubing them yields 64.
4cos 60 i sin 603 64
4cos 180 i sin 1803 64
4cos 240 i sin 240
4cos 330 i sin 330
(b) Since there are four evenly spaced roots on the circle
of radius 4, they are fourth roots of a complex
number of modulus 44. In this case, raising them to
the fourth power yields 128 1283i.
4cos 300 i sin 3003 64
133. z1 2cos i sin z2 2cos i sin z1z2 22cos i sin 4cos i sin 4
z1
2cos i sin z2 2cos i sin 1cos i sin cos2 i sin2 cos 2 cos sin 2 sin isin 2 cos cos 2 sin cos 2 i sin 2
134. (a) z has 4 fourth roots. Three are not shown.
(b) The roots are located on the circle at 30 90k, k 0, 1, 2, 3.
The three roots not shown are located at 120, 210, 300.
606
Chapter 6
Additional Topics in Trigonometry
Problem Solving for Chapter 6
\
1. PQ 2 4.72 62 24.76 cos 25
P
4.7 ft
\
PQ 2.6409 feet
θ
φ
25°
sin sin 25
⇒ 48.78
4.7
2.6409
O
θ
β
6 ft
γ
α
Q
T
180 25 48.78 106.22
180 ⇒ 180 106.22 73.78
106.22 73.78 32.44
180 180 48.78 32.44 98.78
180 180 98.78 81.22
\
PT
4.7
sin 25 sin 81.22
\
PT 2.01 feet
2.
3
mile 1320 yards
4
55°
300 yd
55°
35°
25°
x2 13202 3002 21320300cos 10
θ
1320 yd
x 1025.881 yards 0.58 mile
x
sin sin 10
1320 1025.881
sin 0.2234
180 sin10.2234
167.09
Bearing: 55 90 22.09
S 22.09 E
3. (a)
A
75 mi
30°
15°
135°
x
y
60° Lost party
(c)
B
75°
A
80° 20°
60°
10°
20 mi
Rescue
party
27.452 mi
(b)
x
75
y
75
and
sin 15 sin 135
sin 30 sin 135
x 27.45 miles
y 53.03 miles
z
Lost
party
z2 27.452 202 227.4520 cos 20
z 11.03 miles
sin sin 20
27.45
11.03
sin 0.8511
180 sin10.8511
121.7
To find the bearing, we have 10 90 21.7.
Bearing: S 21.7 E
Problem Solving for Chapter 6
4. (a)
(b)
65°
sin C sin 65
46
52
sin C 46 ft
607
46 sin 65
0.801734
52
C 53.296
52 ft
A 180 B C 61.704
1
(c) Area 4652sin 61.704 1053.09 square feet
2
Number of bags:
a
52
sin 61.704 sin 65
1053.09
21.06
50
a
a 50.52 feet
To entirely cover the courtyard, you would
need to buy 22 bags.
5. If u 0, v 0, and u v 0, then
(a)
u 1, 1,
v 1, 2,
u 2,
(b)
v 5,
u 0, 1,
(c)
(d)
u v 1
u v 3, 2
v 18 32, u v 13
21,
72
u 1,
u uu vv uu vv 1 since all of these are magnitudes of unit vectors.
u v 0, 1
v 3, 3,
u 1,
52 sin 61.704
sin 65
5
2
v 2, 3, u v 3,
v 13, u v ,
u 2, 4,
v 5, 5,
9 494 85
2
u v 7, 1
u 20 25, v 50 52, u v 50 52
6. (a) u 120j
120
(d) tan 40 ⇒ tan1 3 ⇒ 71.565
v 40i
(e)
Up
(b) s u v 40i 120j
140
120
100
Up
80
140
60
120
s
100
80
v
u s
W
E
−20
− 20
20 40 60 80 100
Down
v
20
W
− 60
E
−60
60
40
u
20 40 60 80 100
Down
(c) s 402 1202 16000 4010
126.49 miles per hour
This represents the actual rate of the skydiver’s fall.
s 30i 120j
s 302 1202
15300
123.69 miles per hour
608
Chapter 6
Additional Topics in Trigonometry
8. Let u v 0 and u w 0.
7. Initial point: 0, 0
Terminal point:
u
1
v1 u2 v2
,
2
2
Then, u cv dw u cv u dw
cu v du w
u v1 u2 v2
1
,
w 1
u v
2
2
2
c0 d0
0.
Initial point: u1, u2
Terminal point:
w
u
v
Thus for all scalars c and d, u is orthogonal to cv dw.
1
u v1, u2 v2
2 1
1
v1
u v2
u1, 2
u2
2
2
1
u1 v2 u2
1
,
v u
2
2
2
→
9. W cos F PQ and F1 F2
(a)
F1
If 1 2 then the work is the same since cos cos .
θ1
θ2
F2
P
(b)
Q
If 1 60 then W1 F1
60°
F2
If 2 30 then W2 30°
P
Q
→
1
F PQ
2 1
3
2
→
F2 PQ
W2 3 W1
The amount of work done by F2 is 3 times as great as the amount of work done by F1.
10. (a)
100 sin 100 cos 0.5
0.8727
99.9962
1.0
1.7452
99.9848
1.5
2.6177
99.9657
2.0
3.4899
99.9391
2.5
4.3619
99.9048
3.0
5.2336
99.8630
(b) No, the airplane’s speed does not equal the sum of the vertical and horizontal components of its velocity. To find speed:
speed v sin2 v cos2
(c) (i) speed 5.235 2 149.909 2 150 miles per hour
(ii) speed 10.463 2 149.634 2 150 miles per hour
Practice Test for Chapter 6
Chapter 6
Practice Test
For Exercises 1 and 2, use the Law of Sines to find the remaining sides and
angles of the triangle.
1. A 40, B 12, b 100
2.
C 150, a 5, c 20
3. Find the area of the triangle: a 3, b 6, C 130.
4. Determine the number of solutions to the triangle: a 10, b 35, A 22.5.
For Exercises 5 and 6, use the Law of Cosines to find the remaining sides and
angles of the triangle.
5. a 49, b 53, c 38
6.
C 29, a 100, b 300
7. Use Heron’s Formula to find the area of the triangle: a 4.1, b 6.8, c 5.5.
8. A ship travels 40 miles due east, then adjusts its course 12 southward. After traveling
70 miles in that direction, how far is the ship from its point of departure?
9. w 4u 7v where u 3i j and v i 2j. Find w.
10. Find a unit vector in the direction of v 5i 3j.
11. Find the dot product and the angle between u 6i 5j and v 2i 3j.
12. v is a vector of magnitude 4 making an angle of 30 with the positive x-axis.
Find v in component form.
13. Find the projection of u onto v given u 3, 1 and v 2, 4.
14. Give the trigonometric form of z 5 5i.
15. Give the standard form of z 6cos 225 i sin 225.
16. Multiply 7cos 23 i sin 23 4cos 7 i sin 7.
5
5
i sin
4
4
.
3cos i sin 9 cos
17. Divide
19. Find the cube roots of 8 cos
18. Find 2 2i8.
i sin .
3
3
20. Find all the solutions to x4 i 0.
609