C H A P T E R 6 Additional Topics in Trigonometry Section 6.1
Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . 532
Section 6.2
Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . 539
Section 6.3
Vectors in the Plane
Section 6.4
Vectors and Dot Products
Section 6.5
Trigonometric Form of a Complex Number . . . . . . . . 571
. . . . . . . . . . . . . . . . . . . . 549 . . . . . . . . . . . . . . . . . 562
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606 Practice Test
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609
C H A P T E R 6 Additional Topics in Trigonometry Section 6.1
■
Law of Sines
If ABC is any oblique triangle with sides a, b, and c, then a b c . sin A sin B sin C
■
You should be able to use the Law of Sines to solve an oblique triangle for the remaining three parts, given: (a) Two angles and any side (AAS or ASA) (b) Two sides and an angle opposite one of them (SSA) 1. If A is acute and h b sin A: (a) a < h, no triangle is possible. (b) a h or a > b, one triangle is possible. (c) h < a < b, two triangles are possible. 2. If A is obtuse and h b sin A: (a) a ≤ b, no triangle is possible. (b) a > b, one triangle is possible.
■
The area of any triangle equals one-half the product of the lengths of two sides and the sine of their included angle. 1 1 1 A 2ab sin C 2ac sin B 2bc sin A
Vocabulary Check 1. oblique
2.
1.
b sin B
C b
2.
C 180 A B 105 a 20 sin 45 b 202 28.28 sin B sin A sin 30
532
105°
a 20 sin 105 38.64 sin C sin A sin 30
a 40°
B
Given: A 30, B 45, a 20
c
b
45° c
1 ac sin B 2
C
a = 20
30° A
3.
A
c = 20
B
Given: B 40, C 105, c 20 A 180 B C 35 a
c 20 sin 35 sin A 11.88 sin C sin 105
b
c 20 sin 40 sin B 13.31 sin C sin 105
Section 6.1 3.
4.
C 25° A
C b
a = 3.5
b
a 135° 10°
35° c
Law of Sines
B
A
c = 45
Given: A 25, B 35, a 3.5
Given: B 10, C 135, c 45
C 180 A B 120
A 180 B C 35
b
a 3.5 sin B sin 35 4.75 sin A sin 25
a
c 45 sin 35 sin A 36.50 sin C sin 135
c
3.5 a sin C sin 120 7.17 sin A sin 25
b
45 sin 10 c sin B 11.05 sin C sin 135
5. Given: A 36, a 8, b 5 sin B
b sin A 5 sin 36 0.36737 ⇒ B 21.55 a 8
C 180 A B 180 36 21.55 122.45 c
a 8 sin C sin 122.45 11.49 sin A sin 36
6. Given: A 60, a 9, c 10 sin C
c sin A 10 sin 60 0.9623 ⇒ C 74.21 or C 105.79 a 9
Case 1
Case 2
C 74.21
C 105.79
B 180 A C 45.79
B 180 A C 14.21
b
a 9 sin 45.79 sin B 7.45 sin A sin 60
7. Given: A 102.4, C 16.7, a 21.6 B 180 A C 60.9
b
a 9 sin 14.21 sin B 2.55 sin A sin 60
8. Given: A 24.3, C 54.6, c 2.68 B 180 A C 101.1
b
a 21.6 sin B sin 60.9 19.32 sin A sin 102.4
a
c 2.68 sin 24.3 sin A 1.35 sin C sin 54.6
c
a 21.6 sin C sin 16.7 6.36 sin A sin 102.4
b
c 2.68 sin 101.1 sin B 3.23 sin C sin 54.6
9. Given: A 83 20, C 54.6, c 18.1 B 180 A C 180 83 20 54 36 42 4
10. Given: A 5 40, B 8 15, b 4.8 C 180 A B 166 5
a
c 18.1 sin A sin 83 20 22.05 sin C sin 54.6
a
b 4.8 sin 5 40 sin A 3.30 sin B sin 8 15
b
c 18.1 sin B sin 42 4 14.88 sin C sin 54.6
c
b 4.8 sin 166 5 sin C 8.05 sin B sin 8 15
B
533
534
Chapter 6
Additional Topics in Trigonometry
11. Given: B 15 30 , a 4.5, b 6.8 sin A
a sin B 4.5 sin 15 30 0.17685 ⇒ A 10 11 b 6.8
C 180 A B 180 10 11 15 30 154 19 c
6.8 b sin C sin 154 19 11.03 sin B sin 15 30
12. Given: B 2 45, b 6.2, c 5.8 sin C
c sin B 5.8 sin 2 45 0.04488 ⇒ C 2.57 or 2 34 b 6.2
A 180 B C 174.68, or 174 41 b 6.2 sin 174.68 sin A 11.99 sin B sin 2 45
a
13. Given: C 145, b 4, c 14
14. Given: A 100, a 125, c 10 sin C
b sin C 4 sin 145 0.16388 ⇒ B 9.43 sin B c 14
B 180 A C 75.48
A 180 B C 180 9.43 145 25.57 a
c sin A 10 sin 100 0.07878 ⇒ C 4.52 a 125
b
c 14 sin A sin 25.57 10.53 sin C sin 145
a 125 sin 75.48 sin B 122.87 sin A sin 100
15. Given: A 110 15 , a 48, b 16 sin B
b sin A 16 sin 110 15 0.31273 ⇒ B 18 13 a 48
C 180 A B 180 110 15 18 13 51 32 c
a 48 sin C sin 51 32 40.06 sin A sin 110 15
16. Given: C 85 20, a 35, c 50 sin A
a sin C 35 sin 85 20 0.6977 ⇒ A 44.24, or 44 14 c 50
B 180 A C 50.43, or 50 26 b
C sin B 50 sin 50.43 38.67 sin C sin 85 20
17. Given: A 55, B 42, c
3 4
C 180 A B 83
18. Given: B 28, C 104, a 3 A 180 B C 48
a
c 0.75 sin A sin 55 0.62 sin C sin 83
b
5 a sin B 38 sin 28 2.29 sin A sin 48
b
0.75 c sin B sin 42 0.51 sin C sin 83
c
5 a sin C 38 sin 104 4.73 sin A sin 48
5 8
Section 6.1 19. Given: A 110, a 125, b 100
Law of Sines
535
20. Given: a 125, b 200, A 110
b sin A 100 sin 110 0.75175 ⇒ B 48.74 sin B a 125
No triangle is formed because A is obtuse and a < b.
C 180 A B 21.26 c
a sin C 125 sin 21.26 48.23 sin A sin 110 22. Given: A 76, a 34, b 21
21. Given: a 18, b 20, A 76 h 20 sin 76 19.41
sin B
Since a < h, no triangle is formed.
b sin A 21 sin 76 0.5993 ⇒ B 36.82 a 34
C 180 A B 67.18 c
a sin C 34 sin 67.18 32.30 sin A sin 76
23. Given: A 58, a 11.4, c 12.8 sin B
b sin A 12.8 sin 58 0.9522 ⇒ B 72.21 or B 107.79 a 11.4
Case 1
Case 2
B 72.21
B 107.79
C 180 A B 49.79
C 180 A B 14.21
c
a 11.4 sin 49.79 sin C 10.27 sin A sin 58
24. Given: a 4.5, b 12.8, A 58
c
a 11.4 sin 14.21 sin C 3.30 sin A sin 58
25. Given: A 36, a 5
h 12.8 sin 58 10.86
(a) One solution if b ≤ 5 or b
Since a < h, no triangle is formed. (b) Two solutions if 5 < b < (c) No solution if b >
(a) One solution if b ≤ 10 or b
(c) No solutions if b >
10 . sin 60
10 . sin 60
10 . sin 60
(b) Two solutions if 315.6 < b
5 sin 36
27. Given: A 10, a 10.8
26. Given: A 60, a 10
(b) Two solutions if 10 < b
1 2165
This is not possible. In general, if the sum of any two sides is less than the third side, then they cannot form a triangle. Here 10 5 is less than 16.
54. (a) Working with ODC, we have cos This implies that 2R
a2 . R
a . cos
Since we know that a b c , sin A sin B sin C we can complete the proof by showing that cos sin A. The solution of the system A B C 180
(b) By Heron’s Formula, the area of the triangle is Area ss as bs c. We can also find the area by dividing the area into six triangles and using the fact that the area is 12 the base times the height. Using the figure as given, we have 1 1 1 1 1 1 Area xr xr yr yr zr zr 2 2 2 2 2 2 rx y z rs. rs ss as bs c ⇒
Therefore:
CA
B
r
is 90 A. Therefore: 2R
A
a a a . cos cos90 A sin A
z
B
β α A
y
x D
α O α−C
z
r
β R
s as s bs c.
R
y x
C C
B
Section 6.2
Law of Cosines
547
55. a 25, b 55, c 72 1 (a) Area of triangle: s 25 55 72 76 2
(b) Area of circumscribed circle:
Area 7651214 570.60
cos C
(c) Area of inscribed circle:
51214 7.51 76
r
R
s as bs c s
252 552 722 0.5578 ⇒ C 123.9 22555
1 c 43.37 (see #54) 2 sin C
Area R2 5909.2
(see #54
Area r 2 177.09 56. Given: a 200 ft, b 250 ft, c 325 ft s
200 250 325 387.5 2
Radius of the inscribed circle: r
137.562.5 64.5 ft (see #54) s as s bs c 187.5387.5
Circumference of an inscribed circle: C 2r 264.5 405.2 ft
57.
1 b2 c2 a2 1 bc1 cos A bc 1 2 2 2bc
1 2bc b2 c2 a2 bc 2 2bc
58.
a2 b2 2bc c2 4
1 b c a b c a 4
a2 b c2 4 a b c 2
bca 2
bca 2
abc 2
a b c 2
abc 2
2
60. arccos 0
63. arcsin
3
.
3
2
abc 2
61. arctan3
3
a b c 2
64. arccos
3
3
2
arccos
3
2
5 6 6
66. Let u arccos 3x
2x and 1
1
2
65. Let arcsin 2x, then
1 4x2
2bc a2 b2 c2 1 bc 2 2bc
62. arctan 3 arctan 3
sec
1 b c2 a2 4
59. arcsin1
sin 2x
1 1 a2 b2 c2 bc1 cos A bc 1 2 2 2bc
1 2x
cos u 3x
3x . 1
1
u
θ
3x 1 − 4x 2
tanarccos 3x tan u
1 9x2
3x
1 − 9x 2
548
Chapter 6
Additional Topics in Trigonometry
67. Let arctanx 2, then tan x 2 cot
68. Let u arcsin
x2 and 1
1 . x2
x−2
sin u
θ
1
x1 2
x1 . 2
cos arcsin
2
70. x 2 cos ,
5 25 5 sin 2 5 251
sin2
2 4 4 cos2 2 41 cos2
1 1 cos
2 4 sin2 2 2 sin
csc is undefined.
71. 3 x2 9, x 3 sec
2
2
sin ⇒ cos
3 9
sec2
2
2
sec
1 1 2 cos 22
csc
1 1 2 sin 22
72. x 6 tan ,
3 3 sec 2 9
< < 2 2
12 36 x2
1
12 36 6 tan 2
3 3 tan
12 36 36 tan2
3
12 361 tan2
3
sec 1 tan 2
cot
< < 2 2
2 4 2 cos 2
cos 1
tan
2
2 4 x2
5 5 cos
sec
4 − (x − 1) 2
4 x 12
69. 5 25 x2, x 5 sin
u
x1 cos u 2
1 3 3
2
12 36 sec2
23 3
12 6 sec 2 sec
1 3 tan
cos
csc 1 cot2 1 3 2 2 sin2
12
2
1 2
1
sin2 1 sin ± csc
1 3 4 4
34 ± 23
1 1 2 23 ± ± sin ± 32 3 3
x−1
Section 6.3
5 73. cos cos 2 sin 6 3
5 5 6 3 6 3 sin 2 2
sin x 2 cos 74. sin x 2 2
2 cos
x
x 2 2 2
7 sin 12 4
2 sin
Vectors in the Plane
x
sin
x 2 2 2
2x2 sin 2
2
2 cos x sin
Section 6.3
Vectors in the Plane
\
■
A vector v is the collection of all directed line segments that are equivalent to a given directed line segment PQ .
■
You should be able to geometrically perform the operations of vector addition and scalar multiplication.
■
The component form of the vector with initial point P p1, p2 and terminal point Q q1, q2 is \
PQ q1 p1, q2 p2 v1, v2 v. ■
The magnitude of v v1, v2 is given by v v12 v22.
■
If v 1, v is a unit vector.
■
You should be able to perform the operations of scalar multiplication and vector addition in component form. (a) u v u1 v1, u2 v2
■
(b) ku ku1, ku2
You should know the following properties of vector addition and scalar multiplication. (a) u v v u
(b) u v w u v w
(c) u 0 u
(d) u u 0
(e) cdu cdu
(f) c du cu du
(g) cu v cu cv
(h) 1u u, 0u 0
(i) cv c v v . v
■
A unit vector in the direction of v is u
■
The standard unit vectors are i 1, 0 and j 0, 1. v v1, v2 can be written as v v1i v2 j.
■
A vector v with magnitude v and direction can be written as v ai bj vcos i vsin j, where tan b a.
Vocabulary Check 1. directed line segment
2. initial; terminal
3. magnitude
4. vector
5. standard position
6. unit vector
7. multiplication; addition
8. resultant
9. linear combination; horizontal; vertical
549
550
Chapter 6
Additional Topics in Trigonometry
1. v 4 0, 1 0 4, 1
2. u 3 0, 4 4 3, 8
uv
v 0 3, 5 3 3, 8 uv
3. Initial point: 0, 0
4. Initial point: 0, 0
Terminal point: 3, 2
Terminal point: 4, 2
v 3 0, 2 0 3, 2
v 4 0, 2 0 4, 2
v 32 22 13
v 42 22 20 25
5. Initial point: 2, 2
6. Initial point: 1, 1
Terminal point: 1, 4
Terminal point: 3, 5
v 1 2, 4 2 3, 2
v 3 1, 5 1 4, 6
v 32 22 13
v 42 62 52 213
7. Initial point: 3, 2
8. Initial point: 4, 1
Terminal point: 3, 3
Terminal point: 3, 1
v 3 3, 3 2 0, 5
v 3 4, 1 1 7, 0
v
v 72 02 7
02
52
25 5
9. Initial point: 1, 5
10. Initial point: 1, 11
Terminal point: 15, 12
Terminal point: 9, 3
v 15 1, 12 5 16, 7
v 9 1, 3 11 8, 8
v
162
72
305
v 82 82 82
11. Initial point: 3, 5
12. Initial point: 3, 11
Terminal point: 5, 1
Terminal point: 9, 40
v 5 3, 1 5 8, 6 v
82
62
v 9 3, 40 11 12, 29
100 10
v 122 292 985
13. Initial point: 1, 3
14. Initial point: 2, 7
Terminal point: 8, 9
Terminal point: 5, 17
v 8 1, 9 3 9, 12
v 5 2, 17 7 7, 24
v 9 12 225 15 2
15.
v 72 242 25
2
y
17.
16. 5v
y
y
u+v 5v
v
v x
−v
u x
v x
Section 6.3 18. u v
Vectors in the Plane
551
1 20. v 2 u
19. u 2v y
y
y
u v
u + 2v
x
v − 12 u
2v
u−v
x
−v
− 12 u u
x
21. u 2, 1, v 1, 3 (a) u v 3, 4
(c) 2u 3v 4, 2 3, 9 1, 7
(b) u v 1, 2
y
y
y
5 4
u+v
x
2
v
3
−6
u
1
2
−4
−2
2
4
6
x −3
1
−1
2u
2
3
−2
−1
1
2
3
−6
u x 1
2
3
4
5
−v
−1
2u − 3v
−3v
u−v
− 10
22. u 2, 3, v 4, 0 (b) u v 2, 3
(a) u v 6, 3 y
(c) 2u 3v 4, 6 12, 0 8, 6
y 6
8
y
5 6
10
4
8
2u − 3v
u−v
u
4
u+v
2
2
x 4
6
2
−3v
−v v
2u
4
2
u
x
−5 −4 −3 −2 −1
1
2
− 10 − 8 − 6 − 4 − 2
x 2
4
6
−4
8
−6 −8
23. u 5, 3, v 0, 0 (a) u v 5, 3 u
u=u+v
y
y
7
7
12
6
6
10
5
5
4
4
u=u−v
3
2
1
1 −7 −6 −5 −4 −3 −2 −1
8 6 4
v
x 1
2u = 2u − 3v
3
2 v −7 −6 −5 −4 −3 −2 −1
(c) 2u 3v 2u 10, 6
(b) u v 5, 3 u
y
x 1
2
−3v
− 12 − 10 − 8 − 6 − 4 − 2 −2
2
x
552
Chapter 6
Additional Topics in Trigonometry
24. u 0, 0, v 2, 1 (c) 2u 3v 0, 0 6, 3
(b) u v 2, 1
(a) u v 2, 1
6, 3
y
y
y
1
3
u
2
−3
v=u+v
1
−2
1
2
x
1
−1 −3
− 3v = 2u − 3v
3
−4
−3
−1
2u
−2
−2
x
−1
−7 −6 −5 −4 −3 −2
−1
−v = u − v
u
1
x 1
−5 −6 −7
25. u i j, v 2i 3j
3
5
2
u−v 4
10
−v
x −1
y
2u − 3v 12
u
1 −2
4i 11j
y
y
−3
(c) 2u 3v 2i 2j 6i 9j
(b) u v i 4j
(a) u v 3i 2j
u
u+v
−2
−3
v
−3
8
−3v
3 −1 −2
x
−1
1
2
3
−1
2u x
−8 −6 −4 −2 −2
2
4
6
26. u 2i j, v i 2j (b) u v i j
(a) u v 3i 3j
2u 3v 4i 2j 3i 6j i 4j
y
y 4
u+v
(c)
y
2
3
u
1
3 2 1
2u
2
v
−2
u −4
−3
−2
1
− 6 − 5 −4 − 3
2
x
−1
1 2 3 4
−1
u−v
x
−1
x
−1
−v
1
−2
−1
2u − 3v
− 3v
−5 −6 −7
27. u 2i, v j (c) 2u 3v 4i 3j
(b) u v 2i j
(a) u v 2i j
y
y
y 1
3
1
2u
u 2 1
−1
u+v
v
−1
u −1
1
−1
x 2
x 3
−2
3 −3
−v
u−v
−1
1
2
3
−1 −2 −3 −4
−3v
2u − 2v
x
Section 6.3
Vectors in the Plane
553
28. u 3j, v 2i
y
6i 6j
y
u−v
u+v
3
(c) 2u 3v 6j 6i
(b) u v 2i 3j
(a) u v 2i 3j
y
u
u
2
8
2
2u − 3v 1
1
v −1
1
2
2u 4
−v
x
−3
3
−2
−1
x
−1
1
2
−3v
−1 −8
−6
−4
x
−2
2
−2
29. v
1 1 1 u 3, 0 3, 0 1, 0 u 3 32 02
30. u 0, 2 v
1 1 0, 2 u u 02 22
1 0, 2 0, 1 2
31. u
1 1 1 v 2, 2 2, 2 v 22 22 22
1 1 , 2 2
2 2
33. u
35. u
2
,
2
32. v 5, 12 u
1 1 v 5, 12 52 122 v
1 5, 12 13
135 , 1312
1 1 1 6i 2j v 6i 2j 40 v 62 22
34. v i j
1 1 3 6i 2j j i 10 10 210
u
10 310 i j 10 10
1 1 w 4j j w 4
1 v v 1 12 12
i j
1 2
i j
2
2
i
36. w 6i v
1 1 w 6i w 62 02
1 6i i 6
37. u
1 1 1 i 2j w i 2j w 12 22 5
25 5 2 1 j i i j 5 5 5 5
38. w 7j 3i v
1 1 3i 7j w w 32 72
3 58
i
7 58
j
358 758 i j 58 58
2
2
j
554
Chapter 6
39. 5
u1 u 5 3 1 3 3, 3 35 2 3, 3 2
41. 9
Additional Topics in Trigonometry
40. v 6
2
52, 52 52 2, 52 2
6
u1 u 9 2 1 5 2, 5 929 2, 5 2
u1 u 6 31
2
1829, 4529 182929, 452929
10
43. u 4 3, 5 1
uu 10 0 1
2
u 3i 8j 46. u 0 6, 1 4
45. u 2 1, 3 5 3, 8
u 6, 3
3i 8j
u 6i 3j
48. v 34 w 34 i 2j
47. v 32u j
3i
3,
32
3 4i
3 2j
49. v u 2w
2i j 2i 2j
3 3 4, 2
4i 3j 4, 3
y
y
w
2
y
2w
4
1
1
3 w 4
−1
3
2
x
−1
u + 2w
3
x 1
1
2
2 1
u
−1 x
3u 2
−2
3 −1
51. v 123u w
50. v u w 2i j i 2j i 3j 1, 3
1 2j
7 2,
12
2i j 2i 2j
5j 0, 5 y
−2
2 1
−2
u −2 4 1 (3u + w) 2
−1 −1
1
−2
2
1 w 2
x
x
x
−1
w
−u
5
u
y
3
2
7 2i
4
52. v u 2w
126i 3j i 2j
y
−u + w
44. u 3 0, 6 2
7i 4j
3 2j
1 10, 0 102
1
u 3, 8
1010, 0 10, 0
7, 4
3 2 2i
3, 3
31 2 3, 3 62, 62 32, 32
42. v 10
2
32
3u 2
−3
−2w
−4
u − 2w
3
Section 6.3 53. v 3cos 60i sin 60ºj
54.
v 8cos 135 i sin 135 j
555
55. v 6i 6j
v 8, 135
v 3, 60
Vectors in the Plane
v 62 62 72 62 tan
6 1 6
Since v lies in Quadrant IV, 315. v 5i 4j
56.
57. v 3 cos 0, 3 sin 0
v 5 4 41 2
tan
58. v cos 45, sin 45
3, 0
2
4 5
y
22, 22 y
2
Since v lies in Quadrant II, 141.3.
1
1
x 1
2
3
45° −1
x 1
59. v
72 cos 150, 27 sin 150
73 7 , 4 4
60. v
y
52 cos 45, 25 sin 45
52 52 , 4 4
61. v 32 cos 150, 32 sin 150
36 32 , 2 2
y
y 5
4
3 4
3
3
2
2
2
150° −4
−3
−2
1
1
x
−1
45°
1
1
62. v 43 cos 90, 43 sin 90
0, 43
−5
x
−1
63. v 2
y
2
1 12 32
2 10
6
10
5
4
i 3j
64. v 3
−6
−4
−2
3
10 310 310 j , 5 5 5
1 3i 4j 42
2
y 3
3
x 4
1
9 12 9 12 i j , 5 5 5 5
90° 2
x
−1
3 3i 4j 5
y
2
−2
3
i 3j
i
−3
−1
10 8
−4
150°
6
2
−2 2
1
1
x
−1 −1
x
1
2
1 −1
2
3
556
Chapter 6
Additional Topics in Trigonometry
65. u 5 cos 0, 5 sin 0 5, 0
u 4 cos 60, 4 sin 60 2, 23
66.
v 4 cos 90, 4 sin 90 0, 4
v 5 cos 90, 5 sin 90 0, 5
u v 2, 4 23
u v 5, 5 67. u 20 cos 45, 20 sin 45 102, 102
u 50 cos 30, 50 sin 30 253, 25
68.
43.301, 25
v 50 cos 180, 50 sin 180 50, 0 u v 102 50, 102
v 30 cos 110, 30 sin 110 10.261, 28.191 u v 33.04, 53.19
69. v i j
y
w 2i 2j
1
v
u v w i 3j
x
−1
w 22
1 −1
v w 10 cos
u
α
v 2
v2
2
w
−2
w2
v 2v w
w2
2 8 10 0 22 22
90 70. v i 2j
y
w 2i j
2
u v w i 3j cos
v2
w2
v 2v w
1
w2
5 5 10 0 255
90
Force Two: v 60 cos i 60 sin j Resultant Force: u v 45 60 cos i 60 sin j u v 45 60 cos 2 60 sin 2 90 2025 5400 cos 3600 8100 5400 cos 2475 cos
2475
0.4583 5400
62.7
u
θ −2
−1 −1 −2
71. Force One: u 45i
v
x
w
2
Section 6.3 72. Force One: u 3000i Force Two: v 1000 cos i 1000 sin j Resultant Force: u v 3000 1000 cos i 1000 sin j u v 3000 1000 cos 2 1000 sin 2 3750 9,000,000 6,000,000 cos 1,000,000 14,062, 500 6,000,000 cos 4,062, 500 cos
4,062,500
0.6771 6,000,000
47.4 73.
u 300i v 125 cos 45i 125 sin 45j
R u v 300 R
2
74.
125 125 i j 2 2
300 1252 1252
125 2 tan 125 300 2
125 125 i j 2 2
2
398.32 newtons
⇒ 12.8
u 2000 cos 30 i 2000 sin 30j
y
1732.05i 1000j v 900 cos45i 900 sin45j
2000
636.4i 636.4j u v 2368.4 i 363.6j u v 2368.42 363.62 2396.19 tan
363.6
0.1535 ⇒ 8.7 2368.4
75. u 75 cos 30i 75 sin 30ºj 64.95i 37.5j v 100 cos 45i 100 sin 45j 70.71i 70.71j w 125 cos 120i 125 sin 120j 62.5i 108.3j u v w 73.16i 216.5j u v w 228.5 pounds tan
216.5
2.9593 73.16
71.3º
u+v x
900
Vectors in the Plane
557
558
Chapter 6
Additional Topics in Trigonometry
u 70 cos 30 i 70 sin 30j 60.62i 35j
76.
v 40 cos 45i 40 sin 45j 28.28i 28.28j w 60 cos 135i 60 sin 135j 42.43i 42.43j u v w 46.48i 35.71j u v w 58.61 pounds tan
35.71
0.7683 46.47
37.5 77. Horizontal component of velocity: 70 cos 35 57.34 feet per second Vertical component of velocity: 70 sin 35 40.15 feet per second 78. Horizontal component of velocity: 1200 cos 6 1193.4 ftsec Vertical component of velocity: 1200 sin 6 125.4 ftsec \
79. Cable AC : u ucos 50i sin 50j \
Cable BC : v vcos 30i sin 30j Resultant: u v 2000j u cos 50 v cos 30 0 u sin 50 vsin 30 2000 Solving this system of equations yields: TAC u 1758.8 pounds TBC v 1305.4 pounds \
80. Rope AC : u 10i 24j The vector lies in Quadrant IV and its reference angle is arctan 5 . 12
u u cosarctan
12 5
i sinarctan j 12 5
\
Rope BC : v 20i 24j The vector lies in Quadrant III and its reference angle is arctan5 . 6
v v cosarctan 65 i sinarctan 65 j Resultant: u v 5000j 6 u cosarctan 12 5 v cosarctan 5 0 6 u sinarctan 12 5 v sinarctan 5 5000
Solving this system of equations yields: TAC u 3611.1 pounds TBC v 2169.5 pounds 81. Towline 1: u ucos 18i sin 18j Towline 2: v ucos 18i sin 18j Resultant: u v 6000i u cos 18 u cos 18 6000 u 3154.4 Therefore, the tension on each towline is u 3154.4 pounds.
Section 6.3 82. Rope 1: u u cos 70i sin 70j
70°
Rope 2: v u cos 70i sin 70j
Vectors in the Plane
70°
20° 20°
Resultant: u v 100j u sin 70 u sin 70 100 u 53.2
100 lb
Therefore, the tension of each rope is u 53.2 pounds. y
83. Airspeed: u 875 cos 58i 875 sin 58j
N 140° W
Groundspeed: v 800 cos 50i 800 sin 50j
148°
Wind: w v u 800 cos 50 875 cos 58i 800 sin 50 875 sin 58j
2
138.7 kilometers per hour
Wind speed:
Wind direction: tan Wind direction:
129.2065 50.5507
68.6; 90 21.4
Bearing: N 21.4 E y
84. (a) N W
E S
28° 580 mph
45° x
60 mph
(b) The velocity vector vw of the wind has a magnitude of 60 and a direction angle of 45. vw vwcos i vwsin j 60cos 45i 60sin 45j 60cos 45i sin 45j 60cos 45, sin 45, or 302, 302 (c) The velocity vector vj of the jet has a magnitude of 580 and a direction angle of 118. vj vjcos i vjsin j 580cos 118i 580sin 118j 580cos 118i sin 118j 580cos 118, sin 118 —CONTINUED—
v
40°
Wind speed: w 50.5507 129.2065 2
S x
32°
50.5507i 129.2065j
Wind:
E
u w
559
560
Chapter 6
Additional Topics in Trigonometry
84. —CONTINUED— (d) The velocity of the jet (in the wind) is v vw vj 60cos 45, sin 45 580cos 118, sin 118 60 cos 45 580 cos 118, 60 sin 45 580 sin 118
229.87, 554.54 The resultant speed of the jet is v 229.872 554.542
600.3 miles per hour (e) If is the direction of the flight path, then tan
554.54
2.4124 229.87
Because lies in the Quadrant II, 180 arctan2.4124 180 67.5 112.5. The true bearing of the jet is 112.5 90 22.5 west of north, or 360 22.5 337.5. 85. W FD 100 cos 5030 1928.4 foot–pounds
86. Horizontal force: u ui Weight: w j Rope: t t cos 135i sin 135j
100 lb
u w t 0 ⇒ u t cos 135 0 1 t sin 135 0
50°
t 2 pounds
30 ft
u 1 pound 87. True. See Example 1.
88. True. u a2 b2 1 ⇒ a2 b2 1
89. (a) The angle between them is 0. (b) The angle between them is 180. (c) No. At most it can be equal to the sum when the angle between them is 0. 90. F1 10, 0, F2 5cos , sin (a)
F1 F2 10 5 cos , 5 sin
(b)
15
F1 F2 10 5 cos 2 5 sin 2 100 100 cos 25 cos2 25 sin2 54 4 cos cos2 sin2
2
0 0
54 4 cos 1 55 4 cos (c) Range: 5, 15 Maximum is 15 when 0. Minimum is 5 when . (d) The magnitude of the resultant is never 0 because the magnitudes of F1 and F2 are not the same.
Section 6.3
Vectors in the Plane
91. Let v cos i sin j. v cos2 sin2 1 1 Therefore, v is a unit vector for any value of . 92. The following program is written for a TI-82 or TI-83 or TI-83 Plus graphing calculator. The program sketches two vectors u ai bj and v ci dj in standard position, and then sketches the vector difference u v using the parallelogram law.
93. u 5 1, 2 6 4, 4
PROGRAM: SUBVECT :Input “ENTER A”, A :Input “ENTER B”, B :Input “ENTER C”, C :Input “ENTER D”, D :Line (0, 0, A, B) :Line (0, 0, C, D) :Pause :A – C→E :B – D→F :Line (A, B, C, D) :Line (A, B, E, F) :Line (0, 0, E, F) :Pause :ClrDraw :Stop 94.
v 9 4, 4 5 5, 1
u 80 10, 80 60 70, 20 v 20 100, 70 0 80, 70
u v 1, 3 or v u 1, 3
u v 70 80, 20 70 10, 50 v u 80 70, 70 20 10, 50 96. x 8 sin
95. x2 64 8 sec 2 64
64 x2 64 8 sin2
64sec2 1
64 64 sin2
8tan2
8 tan for 0 < < 2
81 sin2 8cos2 8 cos for 0 <
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 km hr 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
v 0
⇒ Not orthogonal
Neither 88. u 2i j, v 3i 6j u
v0
⇒ Orthogonal
89. u 4, 3, v 8, 2 w1 projvu
v v 68 8, 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 v v 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
52 1, 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
