Precalculus Mathematics For Calculus International Metric Edition 7th Edition Stewart Solutions Manual Instant Download All Chapter

Precalculus Mathematics for Calculus
International Metric Edition 7th Edition

Stewart Solutions Manual
Go to download the full and correct content document:
https://testbankdeal.com/product/precalculus-mathematics-for-calculus-international-
metric-edition-7th-edition-stewart-solutions-manual/
More products digital (pdf, epub, mobi) instant
download maybe you interests ...
Precalculus Mathematics for Calculus 7th Edition

Stewart Solutions Manual
https://testbankdeal.com/product/precalculus-mathematics-for-
calculus-7th-edition-stewart-solutions-manual/

Stewart Test Bank
calculus-7th-edition-stewart-test-bank/

Stewart Test Bank
calculus-6th-edition-stewart-test-bank/
Calculus 7th Edition Stewart Solutions Manual
https://testbankdeal.com/product/calculus-7th-edition-stewart-
solutions-manual/
Multivariable Calculus 7th Edition Stewart Solutions
Manual
https://testbankdeal.com/product/multivariable-calculus-7th-
edition-stewart-solutions-manual/
Calculus An Applied Approach Brief International Metric

Edition 10th Edition Larson Solutions Manual
https://testbankdeal.com/product/calculus-an-applied-approach-
brief-international-metric-edition-10th-edition-larson-solutions-
manual/
Calculus Early Transcendentals 7th Edition Stewart

Solutions Manual
https://testbankdeal.com/product/calculus-early-
transcendentals-7th-edition-stewart-solutions-manual/
Understanding Basic Statistics International Metric

Edition 7th Edition Brase Solutions Manual
https://testbankdeal.com/product/understanding-basic-statistics-
international-metric-edition-7th-edition-brase-solutions-manual/
Calculus 7th Edition Stewart Test Bank
https://testbankdeal.com/product/calculus-7th-edition-stewart-
test-bank/
6 TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH
6.1 ANGLE MEASURE

1. (a) The radian measure of an angle is the length of the arc that subtends the angle in a circle of radius 1.
.
(b) To convert degrees to radians we multiply by 180
(c) To convert radians to degrees we multiply by 180
.
2. (a) If a central angle is drawn in a circle of radius r, the length of the arc subtended by is s r .
(b) The area of the sector with central angle is A 12 r 2 .

3. (a) The angular speed of the point is $ .
t
s
(b) The linear speed of the point is ) .
t
(c) The linear speed ) and the angular speed ome are related by the equation ) r$.
4. No, if the common angular speed is $, Object A has linear speed 2$, while Object B has angular speed 5$. Object B has
greater linear speed.
rad rad s 0262 rad
5. 15i 15i 180 rad rad s 0628 rad
6. 36i 36i 180
i 12 i 5
rad 3 rad s 0942 rad
7. 54i 54i 180 rad 5 rad s 1309 rad
8. 75i 75i 180
i 10 i 12
rad rad s 0785 rad
9. 45i 45i 180 rad rad s 0524 rad
10. 30i 30i 180
i 4 i 6
11. 100i 100i 180 rad 10 rad s 3491 rad
12. 200i 200i 180
i 9 i 9
13. 1000i 1000i 180 rad 20 rad s 62832 rad
14. 3600i 3600i 180
i 9 i

15. 70i 70i 180 rad 5 rad s 2618 rad
16. 150i 150i 180
i 18 i 6
i i
17. 53 53 180
300
i 18. 34 34 180
135
i
i i
19. 56 56 180
150
i 20. 32 32 180
270
i
i 540i i 360i
21. 3 3 180
s 1719
i 22. 2 2 180
s 1146
i
i 216i i 612i
23. 12 12 180
688
i 24. 34 34 180
s 1948
i
180i 18i
25. 10 5 180i 50i
26. 518
10 18
i
2 180 24i 13 180 i
27. 215 15 28. 1312 12 195
i
29. 50i is coterminal with 50i 360i 410i , 50i 720i 770i , 50i 360i 310i , and 50i 720i 670i . (Other
answers are possible.)
30. 135i is coterminal with 135i 360i 495i , 135i 720i 855i , 135i 360i 225i , and 135i 720i 585i .
(Other answers are possible.)
31. 34 is coterminal with 34 2 114 , 34 4 194 , 34 2 54 , and 34 4 134 . (Other answers are
possible.)
32. 116 is coterminal with 116 2 236 , 116 4 356 , 116 2 11 13
6 , and 6 4 6 . (Other answers
are possible.)
477
478 CHAPTER 6 Trigonometric Functions: Right Triangle Approach
33. 7 15 9 17

4 is coterminal with 4 2 4 , 4 4 4 , 4 2 4 , and 4 4 4 . (Other answers
are possible.)
34. 45i is coterminal with 45i 360i 315i , 45i 720i 675i , 45i 360i 405i , and 45i 720i 765i .
(Other answers are possible.)
35. Since 430i 70i 360i , the angles are coterminal.
36. Since 330i 30i 360i , the angles are coterminal.
37. Since 176 56 126 2; the angles are coterminal.
38. Since 323 113 213 7 is not a multiple of 2, the angles are not coterminal.
39. Since 875i 155i 720i 2 360i , the angles are coterminal.
40. Since 340i 50i 290i is not a multiple of 360i , the angles are not coterminal.
41. Since 400i 360i 40i , the angles 400i and 40i are coterminal.
42. Since 375i 360i 15i , the angles 375i and 15i are coterminal.
43. Since 780i 2 360i 60i , the angles 780i and 60i are coterminal.
44. Since 100i 260i 360i is a multiple of 360i , the angles 100i and 260i are coterminal.
45. Since 800i 3 360i 280i , the angles 800i and 280i are coterminal.
46. Since 1270i 190i 1080i 3 360i is a multiple of 360i , the angles 1270i and 190i are coterminal.
47. Since 196 2 76 , the angles 196 and 76 are coterminal.
48. Since 53 2 5
3 , the angles 3 and 3 are coterminal.
49. Since 25 12 2 , the angles 25 and are coterminal.
50. Since 10 2 s 3717, the angles 10 and 10 2 are coterminal.
51. Since 174 2 2 17
4 , the angles 4 and 4 are coterminal.
52. Since 512 32 24 12 2, the angles 512 and 32 are coterminal.
53. Using the formula s r, s 56 9 152 .
r s 7 35
54. Using the formula s r, the length of the arc is s 140i 5 5 s 122.
180i 9 9
s 10 180 i
55. 2 rad 2 s 1146i
r 5
s 8
56. Solving for r, we have r , so the radius of the circle is r 4.
2
57. Solving for s, we have s r , so the length of the arc is 5 3 15 cm.
8
58. Solving for s, we have s r 12 40i s 838 m.
180i 3
s 14 14 180 i
59. Solving for , we have rad s 891i .
r 9 9
s 5 5 180i
60. Solving for , we have rad s 955i .
r 3 3
s 15 18
61. r s 573 m
56
s 20 72
62. r i i 2292 cm
50 180
32 4 128 s 4468
63. (a) A 12 r 2 12 82 80i 180i 9 9
(b) A 12 r 2 12 102 05 25
U U
2A 2 12
64. (a) A 12 r 2 " r , so the radius is s 586.
07
SECTION 6.1 Angle Measure 479
U V
2A 2 12
(b) r s 303
150i 180i
r s
65. 23 rad and r 10 m, so A 12 r 2 12 102 23 100
3 s 1047 m .
2
r s
66. 145i 145i 29 1 2 1 2 29 29 s 51 m 2.
180i 36 and r 2 m , so A 2 r 2 2 36 18
7 1 7 2
67. A 70 m2 and 140i 140i rad. Thus, A 12 r 2 % 70 r "
180i 9 2 9
U T
9 6 5
r 2 70 s 76 m.
7
U T
68. A 20 m2 and 512 , so A 1 r 2 % 20 5 r 2 " r 2 20 12 4 6 s 55 m.
2 12 5
69. r 80 km and A 1600 km2, so A 12 r 2 % 1600 12 802 % 12 rad.
U
2 2 600
70. The area of the circle is r 600 m , so r . Thus, the area of the sector is

1 600 900
A 12 r 2 3 s 2865 m2 .
2
71. Referring to the figure, we have AC 3 1 4,
BC 1 2 3, and AB 2 3 5. Since
$
AB 2 AC 2 BC 2 , then by the Pythagorean Theorem, the
triangle is a right triangle. Therefore,

2 and

#
A 12 r 2 12 12 2
2 4 m .
%
s 1
72. The triangle is equilateral, so 1
3 rad. To find 2 , we use the formula 2 r 1 1 rad. Thus,
1 2 i
3 1 s 0047 rad, or approximately 27 .
73. Between 1:00 P. M . and 1:00 P. M ., the minute hand traverses three-quarters of a complete revolution, or 34 2 32 rad,
while the hour hand moves three-quarters of the way from 12 to 1, which is itself one-twelfth of a revolution. So the hour
r s
1 2 rad.
hand traverses 34 12 8
74. Between 1:00 P. M . and 6:45 P. M ., the minute hand traverses five complete revolutions plus three-quarters of a revolution;
that is, 5 2 34 2 232 rad, while the hour hand moves through five-twelfths of a revolution, plus three-quarters of
r s
5 2 3 1 2 23 rad.
the way from 6 to 7; that is, 12 4 12 24
75. The circumference of each wheel is d 70 cm. If the wheels revolve 10,000 times, the distance traveled is
1m 1 km
10,000 70 cm s 21 99 k m.
100 cm 1000 m
76. Since the diameter is 75 cm, we have r 37. 5 cm . In one revolution, the arc length (distance traveled) is s r 2
37. 5 75 c m . The total distance traveled is 1 k m 1000 m/k m 100 cm/m 100,000 cm 100 ,00 0 cm
1 rev
s 424 41 rev. Therefore the car wheel will make approximately 424 revolutions.
75 cm
rad rad.
77. We find the measure of the angle in degrees and then convert to radians. 405i 255i 15i and 15 180i 12
Then using the formula s r , we have s
12 6340 528 s 1659808 a nd so the distance between the two cities is
roughly 1660 k m.
78. 35i 30i 5i 5i 180 rad rad. Then using the formula s r , the length of the arc is
i 36
s
36 6340 176 s 55327. So the distance between the two cities is roughly 553 km.
1 of its orbit which is 2 rad. Then s r 2 149 ,000,000 s 2 ,564 ,916 7 , so the
79. In one day, the earth travels 365 365 365
distance traveled is approximately 25 6 million kilometers.
80. Since the sun is so far away, we can assume that the rays of the sun are parallel when striking the earth. Thus, the angle
s 800 180 8 00
formed at the center of the earth is also 72i . So r 72 s 6 368 k m, and the circumference
72i 180i
2 180 8 00
is c 2r 40 ,000 k m.
72
r s
81. The central angle is 1 minute 60 1 i 1 rad rad. Then s r 6 34 0 s 1844 , and so a
60 180i 10,800 10,800
nautical mile is approximately 1844 k m.
b c s 19,791 m2 .
82. The area is A 12 r 2 12 902 280i 180i
83. The area is equal to the area of the large sector (with radius 85 cm ) minus the area of the small sector (with radius 35 cm )
r sb
Thus, A 12 r12 12 r22 12 852 35 2 135i 180 c s 7069 cm 2 .
i
84. The area available to the cow is shown in the diagram. Its area is
the sum of four quarter-circles:
m
r s
A 14 302 152 122 92
m
m m
337.5 m
s 1060 m2
m
m
m
45 2 rad
85. (a) The angular speed is $ 90 rad/min.
1 min
45 2 40
(b) The linear speed is ) 3600 cm /min s 11,309 7 cm /min
1
1000 2 rad
1 min
1000 2 115
00 5 m /s s 15 7 m /s.
(b) The linear speed is )
60
8 2 1 16
87. ) s 3.351 m/s.
15 15
600 2 128
00 m 1 km 60 min 20 16 k m/h s 6 33 k m/h.
88. )
1 min 1000 m 1 hr
1 2 6340 1 day
89. 23 h 56 min 4 s 239344 hr. So the linear speed is s 166436 k m/h.
1 day 239344 hr
linear speed 8 0 k m/h 1 h 1000 m
90. (a) The radius is 1. 2 m , so the angular speed is $ 1111 rad/min
.
radius 1. 2 m 60 min 1 k m
angular speed 1111 rad/min
(b) The rate of revolution is s 177 rev/min.
2 2
100 2 020 m 2
91. ) s 209 m/s.
60 s 3
linear speed of pedal 40 2 10
radius of wheel sprocket 5
(b) The linear speed of the bicycle is ) angular speed radius 160 rad/min 32 .5 cm 5 200 cm /min s 9 8 k m/h.
SECTION 6.2 Trigonometry of Right Triangles 481
93. (a) The circumference of the opening is the length of the arc subtended by the angle on the flat piece of paper, that is,
C s r 6 53 10 s 314 cm.
C 10
(b) Solving for r, we find r 5 cm.
2 2
T
(c) By the Pythagorean Theorem, h 2 62 52 11, so h 11 s 33 cm.
T
(d) The volume of a cone is V 13 r 2 h. In this case V 13 52 11 s 868 cm3 .
94. (a) With an arbitrary angle , the circumference of the opening is (b)
V 100
C 3 S 92
C 6, r 2 2
, h 6 r 36 2 , and
2 50
V
92 9 2 9 S
V 13 r 2 h 2 36 2 2 2 42 2 . 0
3 0 2 4 6
(c) The volume seems to be maximized for s 513 rad or about 293i .
95. Answers will vary, although of course everybody prefers radians.
6.2 TRIGONOMETRY OF RIGHT TRIANGLES

opposite adjacent opposite
1. (a) opposite (b) sin , cos , and tan .
hypotenuse hypotenuse adjacent
adjacent (c) The trigonometric ratios do not depend on the size of the triangle because
all right triangles with angle are similar.
hypotenuse n
1 1 1
2. The reciprocal identities state that csc , sec , and cot .
sin cos tan
3. sin 45 , cos 35 , tan 43 , csc 54 , sec 53 , cot 34
7 , cos 24 , tan 7 , csc 25 , sec 25 , cot 24
4. sin 25 25 24 7 24 7
S T
5. The remaining side is obtained by the Pythagorean Theorem: 41 402 81 9. Then sin 40
2 9
41 , cos 41 ,
tan 40 41 41 9
9 , csc 40 , sec 9 , cot 40
S T
6. The hypotenuse is obtained by the Pythagorean Theorem: 82 152 289 17. Then sin 15 8
17 , cos 17 ,
tan 15 17 17 8
8 , csc 15 , sec 8 , cot 15
S T T
7. The remaining side is obtained by the Pythagorean Theorem: 32 22 13. Then sin T2 2 1313 ,
13
T T T
cos T3 3 1313 , tan 23 , csc 213 , sec 313 , cot 32
13
S T T
8. The remaining side is obtained by the Pythagorean Theorem: 82 72 15. Then sin 78 , cos 815 ,
T T T
tan T7 7 1515 , csc 87 , sec T8 8 1515 , cot 715
15 15
S T
2 2
9. c 5 3 34
T T
(a) sin cos T3 3 3434 (b) tan cot 35 (c) sec csc 534
34
S T
2 2
10. b 7 4 33
(a) sin cos 47 (b) tan cot T4 (c) sec csc T7

33 33
11. (a) sin 22i s 037461 (b) cot 23i s 235585
12. (a) cos 37i s 079864 (b) csc 48i s 134563
13. (a) sec 13i s 102630 (b) tan 51i s 123490
14. (a) csc 10i s 575877 (b) sin 46i s 071934

x
15. Since sin 30i , we have x 25 sin 30i 25 12 25
2.
25
12 12 12 T
16. Since sin 45i , we have x i 1 12 2.
x sin 45 T
2
x T T
17. Since sin 60i , we have x 13 sin 60i 13 23 132 3 .
13
4 4 4 T
18. Since tan 30i , we have x 1 4 3.
x tan 30i T
3
12 12
19. Since tan 36i , we have x s 1651658.
x tan 36i
25 25
20. Since sin 53i , we have x s 3130339.
x sin 53i
x y
21. cos % x 28 cos , and sin % y 28 sin .
28 28
x 4 4
22. tan % x 4 tan , and cos % y 4 sec .
4 y cos
S T
23. tan 56 . Then the third side is x 52 62 61. 24. cos 12
13 . The third side is
T T S T
The other five ratios are sin 5 6161 , cos 6 6161 , y 132 122 25 5. The other five ratios are
T T 5 , tan 5 , csc 13 , sec 13 , and
sin 13
csc 561 , sec 661 , and cot 65 . 12 5 12
cot 12
5.

n
n

S T T S
25. cot 1. Then the third side is r 12 12 2. 26. tan 3. The third side is r 12 3 2. The other
T T
The other five ratios are sin T1 22 , five ratios are sin 23 , cos 12 , csc T2 ,
2 3
T T
cos T1 22 , tan 1, csc 2, and sec 2, and cot T1 .
2 3
T
sec 2. n

n

S T S T
27. csc 11
6 . The third side is x 112 62 85. The 28. cot 53 . The third side is x 52 32 34. The
T T T
6 , cos 85 ,
other five ratios are sin 11 other five ratios are sin 3 3434 , cos 5 3434 ,
11
T T T T T
tan 6 8585 , sec 118585 , and cot 685 . tan 35 , csc 334 and sec 534 .

n n

T T
29. sin
6 cos 1 3 1 3
6 2 2 2
1
30. sin 30i csc 30i sin 30i 1
sin 30i
T T
31. sin 30i cos 60i sin 60i cos 30i 12 12 23 23 14 34 1
r T s2 r s2
32. sin 60i 2 cos 60i 2 23 12 34 14 1
r T s2
r s2
33. cos 30i 2 sin 30i 2 3
12 34 1 1
2 4 2
b rT s2 K rT sL2 rT s2 r T s
34. sin cos sin cos c2 3 T1 T1 12 T 1 31 18 3 1 18 3 2 3 1
3 4 4 3 r2 2T s 2 r T 2s 2
18 4 2 3 1 2 3
4
b rT s2 r srT s T
35. cos sin c2 2 1 1 1 2 1 2 3 2
4 6 2 2 2 4 2 2 4 2
b c rT T T s2 r T s2 T
36. sin 3 tan 2 3 3 12 2 94 2
6 csc 4 2 3 2
37. This is an isosceles right triangle, so the other leg has length 16 tan 45i 16, the hypotenuse has length
16 T
i 16 2 s 2263, and the other angle is 90i 45i 45i .
sin 45
100
38. The other leg has length 100 tan 75i s 2679, the hypotenuse has length s 10352, and the other angle is
sin 75i
i i
90 75 15 . i
35
39. The other leg has length 35 tan 52i s 4480, the hypotenuse has length s 5685, and the other angle is
cos 52i
i i
90 52 38 . i
40. The adjacent leg has length 1000 cos 68i s 37461, the opposite leg has length 1000 sin 68i s 92718, and the other angle
is 90i 68i 22i .
41. The adjacent leg has length 335 cos
8 s 3095, the opposite leg has length 335 sin 8 s 1282, and the other angle is
3 .
2 8 8
723
42. The opposite leg has length 723 tan
6 s 4174, the hypotenuse has length s 8348, and the other angle is
cos
6
.
2 6 3
106 106
43. The adjacent leg has length s 14590, the hypotenuse has length s 18034, and the other angle is
tan
5 sin 5
3 .
2 5 10
44. The adjacent leg has length 425 cos 38 s 16264, the opposite leg has length 425 sin 38 s 39265, and the other angle is
3 .
2 8 8
1 s 045 cos s 2 s 089, tan 1 , csc s 224, sec s 224 s 112, cot s 200.
45. sin s 224 224 2 2
064
46. sin 40i s 064, cos 40i s 077, tan 40i s s 083, csc 40i s 156,
077
sec 40i s 131, cot 40i s 120.
c
100 100
47. x s 2309
tan 60i tan 30i
85 85 85
48. Let d be the length of the base of the 60i triangle. Then tan 60i %d s 49075, and so tan 30i
d tan 60i dx
85 85
%dx %x d s 981.
tan 30i tan 30i
50 50 h h
49. Let h be the length of the shared side. Then sin 60i %h s 57735 % sin 65i % x s 637
h sin 60i x sin 65i
5 5 h
50. Let h be the hypotenuse of the top triangle. Then sin 30i %h 10, and so tan 60i %
h sin 30i x
h 10
x s 58.
tan 60i tan 60i
x y
51. From the diagram, sin and tan , so x y sin 10 sin tan .
y 10
n
Z
y Y
n

a b 1 d
52. sin % a sin , tan % b tan , cos % c sec , cos % d cos
1 1 c 1
i h
53. Let h be the height, in meters , of the Empire State Building. Then tan 11 % h 16 00 tan 11i s 311 m .
1600
10,500
54. (a) Let r be the distance, in meters , between the plane and the Gateway Arch. Therefore, sin 22i %
r
10,500
r s 28 ,029 m .
sin 22i
(b) Let x be the distance, also in meters , between a point on the ground directly below the plane and the Gateway Arch. Then
10,500 10,500
tan 22i %x s 25,988 m.
x tan 22i
h
55. (a) Let h be the distance, in kilometers, that the beam has diverged. Then tan 05i %
384,000
h 384 ,000 tan 05i s 3351 k m.
(b) Since the deflection is about 3351 k m whereas the radius of the moon is about 16 00 k m, the beam will not strike the
moon.
60 60
56. Let x be the distance, in meters , of the ship from the base of the lighthouse. Then tan 23i %x s 1 41 m .
x tan 23i
h
57. Let h represent the height, in meters , that the ladder reaches on the building. Then sin 72i % h 6 sin 72i s 5.7 m .
6
h
58. Let h be the height, in meters, of the communication tower. Then sin 65i % h 180 sin 65i s 163 m .
180
h
59. Let h be the height, in meters , of the kite above the ground. Then sin 50i % h 135 sin 50i s 103 m.
135
60. Let h 1 be the height of the flagpole above elevation and let h 2 be the height below,
I|
h
c Y as shown in the figure. So tan 18i 1 % h 1 x tan 18i . Similarly,
c x
Il
h 2 x tan 14i . Since the flagpole is 18 meters tall, we have h 1 h 2 18 , so
18
x tan 18i tan 14i 18 % x s 3 14 m .
tan 18i tan 14i
h
61. Let h 1 be the height of the window in meters and h 2 be the height from the window to the top of the tower. Then tan 25i 1
98
i i h2 i
% h 1 98 tan 25 s 46 m . Also, tan 39 % h 2 98 tan 39 s 79 m . Therefore, the height of the window
98
is approximately 46 m and the height of the tower is approximately 46 79 12 5 m .
62. c c Let d1 be the distance, in meters , between a point directly below the plane and one
car, and d2 be the distance, in meters , between the same point and the other car. Then
c c 1545 1545 1545
E| El tan 52i % d1 s 1207 09, and tan 35i %
d1 tan 52i d2
1545
d2 s 2206 49 m. So the distance between the two cars is about d1 d2 s 1207 0 9 2206 49 s 3414 m .
tan 35i
63. Let d1 be the distance, in meters , between a point directly below the plane and one car, and d2 be the distance, in meters , between
d d
the same point and the other car. Then tan 52i 1 % d1 1 54 5 tan 52 i s 1977 5 m. Also, tan 38 i 2 %
1545 1545
d2 1 54 5 tan 38 i s 1207 09 m . So in this case, the distance between the two cars is about 77 0 m .
64. Let x be the distance, in meters , between a point directly below the balloon and the first kilometer post. Let h be the height,
h h
in meters , of the balloon. Then tan 22i and tan 20i . So h x tan 22i x 1000 tan 20i %
x x 1000
1000 tan 20i
x s 9087 m . Therefore h s 9087 tan 22 i s 3671 m s 37 k m.
tan 22i tan 20i
65. Let x be the horizontal distance, in meters , between a point on the ground directly below the top of the mountain and
h
the point on the plain closest to the mountain. Let h be the height, in meters , of the mountain. Then tan 35i
x
i h i i 1000 tan 32i
and tan 32 . So h x tan 35 x 1000 tan 32 % x s 82942. Thus
x 1000 tan 35i tan 32i
h s 82942 tan 35i s 5808 m .
66. Since the angle of elevation from the observer is 45i , the distance from the
observer is h, as shown in the figure. Thus, the length of the leg in the smaller right
I h
triangle is 600 h. Then tan 75i % 600 h tan 75i h %
600 h
c 600 tan 75i
600 tan 75i h 1 tan 75i % h s 473 m.
I

I 1 tan 75i
d
67. Let d be the distance, in kilometers, from the earth to the sun. Then sec 8985i %
384,000
d 384 ,000 sec 8985 i s 146 7 million kilometers.
68. (a) s r % rs 9848
6340 s 1553 3 rad s 89
i
6340
(b) Let d represent the distance, in kilometers , from the center of the earth to the moon. Since cos , we have
d
6340 6340
d s s 363, 2737. So the distance AC is 363, 2737 6 34 0 s 356,900 km .
cos cos 89 i
69. Let r represent the radius, in kilometers, of the earth. Then sin 60276i rr960 % r 9 60 sin 60276i r %
60 sin 60276i s 6 33 6158 . So the earth’s radius is about 6 34 0 k m.
960 sin 60276i r 1 sin 60276i % r 91sin 60276i
70. Let d represent the distance, in kilo meters , from the earth to Alpha Centauri. Since sin 0000211i 149,000 d
,000
, we have
149, 000,000 13
d sin 0000211i s 40 ,46 5,283 ,160 ,456 . So the distance from the earth to Alpha Centauri is about 40 5 10 km.
d
71. Let d be the distance, in AU, between Venus and the sun. Then sin 463i d, so d sin 463i s 0723 AU.
1
72. If two triangles are similar, then their corresponding angles are equal and their corresponding sides are proportional. That
n n n n n n
is, if triangle ABC is similar to triangle A) B ) C ) then AB r n A) B ) n, AC r n A) C ) n, and BC r n B ) C ) n. Thus when
we express any trigonometric ratio of these lengths as a fraction, the factor r cancels out.
6.3 TRIGONOMETRIC FUNCTIONS OF ANGLES

1. If the angle is in standard position and P x y is a point on the terminal side of , and r is the distance from the origin to
y x y
P, then sin , cos , and tan .
r r x
2. The sign of a trigonometric function of depends on the quadrant in which the terminal angle of lies. For example, if
is in quadrant II, sin is positive. In quadrant III, cos is negative. In quadrant IV, sin is negative.
3. (a) If is in standard position, then the reference angle is the acute angle formed by the terminal side of and the x-axis.
So the reference angle for 100i is 80i and that for 190i is 10i .
(b) If is any angle, the value of a trigonometric function of is the same, except possibly for sign, as the value of the
trigonometric function of . So sin 100i sin 80i and sin 190i sin 10i .
4. The area A of a triangle with sides of lengths a and b and with included angle is given by the formula A 12 ab sin . So
the area of the triangle with sides 4 and 7 and included angle 30i is 12 4 7 sin 30i 7.
5. (a) The reference angle for 120i is 180i 120i 60i . 6. (a) The reference angle for 175i is 180i 175i 5i .
(b) The reference angle for 200i is 200i 180i 20i . (b) The reference angle for 310i is 360i 310i 50i .
(c) The reference angle for 285i is 360i 285i 75i . (c) The reference angle for 730i is 730i 720i 10i .
7. (a) The reference angle for 225i is 225i 180i 45i . 8. (a) The reference angle for 99i is 180i 99i 81i .
(b) The reference angle for 810i is 810i 720i 90i . (b) The reference angle for 199i is
199i 180i 19i .
(c) The reference angle for 105i is
180i 105i 75i . (c) The reference angle for 359i is 360i 359i 1i .
is 7 3 .
9. (a) The reference angle for 710 10. (a) The reference angle for 56 is 56
10 10 6.
(b) The reference angle for 98 is 98

8. (b) The reference angle for 109 is 109
9.
(c) The reference angle for 103 is 103 3

3. (c) The reference angle for 237 is 237 3 27 .
11. (a) The reference angle for 57 is 57 27 . 12. (a) The reference angle for 23 is 23 2 03.
(b) The reference angle for 14 is 14 04. (b) The reference angle for 23 is 23 s 084.
(c) The reference angle for 14 is 14 because 14

2. (c) The reference angle for 10 is 10 10 0.
T T
13. cos 150i cos 30i 23 14. sin 240i sin 60i 23
SECTION 6.3 Trigonometric Functions of Angles 487
T
15. tan 330i tan 30i 33 16. sin 30i sin 30i 12
T T
17. cot 120i cot 60i 33 18. csc 300i csc 60i 2 3 3
T
19. csc 630i csc 90i sin190i 1 20. cot 210i cot 30i tan130i 3
T
21. cos 570i cos 30i 23 22. sec 120i sec 60i cos160i 2
T
23. tan 750i tan 30i T1 33 24. cos 660i cos 60i 12
3
25. sin 32 sin

2 1 26. cos 43 cos 1
3 2
r s T r s T
27. tan 43 tan 3 3 28. cos 116 cos 6 2
3
r s b c T
29. csc 56 csc 6 2 30. sec 76 sec 2 3
6 3
1 1 T
31. sec 173 sec
3 cos 2 32. csc 54 csc
4 2
3 sin
4
b c 1 T
33. cot 4 cot 4 tan 1 34. cos 74 cos T1
4 2 2
2
4
35. tan 52 tan 2 which is undefined. 36. sin 116 sin 6 2
1
37. Since sin 0 and cos 0, is in quadrant III.

38. Since both tan and sin are negative, is in quadrant IV.
sin
39. sec 0 " cos 0. Also tan 0 " 0 % sin 0 (since cos 0). Since sin 0 and cos 0, is
cos
in quadrant IV.
40. Since csc 0 " sin 0 and cos 0, is in quadrant II.
S
S sin 1 cos2
2
41. Since sin is negative in quadrant III, sin 1 cos and we have tan .
cos cos
S
cos 1 sin2
42. cot because cos 0 in quadrant II.
sin sin
S
43. cos2 sin2 1 % cos 1 sin2 because cos 0 in quadrant IV.
1 1
44. sec S because all trigonometric functions are positive in quadrant I.
cos 1 sin2
S
45. sec2 1 tan2 % sec 1 tan2 because sec 0 in quadrant II.
S
46. csc2 1 cot2 % csc 1 cot2 because csc 0 in quadrant III.
S T
47. sin 45 . Then x 52 42 9 3, since is in quadrant IV. Thus, cos 35 , tan 43 , csc 54 ,
sec 53 , and cot 34 .
S
48. tan 43 , so r 42 32 5. Since is in quadrant III, x 3 and y 4, and so sin 45 , cos 35 ,
csc 54 , sec 53 , and cot 34 .
U r s2 T T T T
7
49. cos 12 , so sin 1 12 7 1295 , tan 795 , csc 129595 , sec 12
7 , and cot 7 95
95
S T T T T
8
50. cot 9 , so r 82 92 145. Thus, sin T 9 145 , cos 145 , tan 8 , csc 145
9 145 8 145 9
9 ,
145
T
and sec 145
8 .
S T T T T
51. csc 2. Then sin 12 and x 22 12 3. So sin 12 , cos 23 , tan T1 33 , sec T2 2 3 3 ,
3 3
T
and cot 3.
S T T
52. cot 14 . Then tan 4 and r 42 12 17. So sin T4 , cos T1 , tan 4, csc 417 , and
17 17
T
sec 17.
S T T T T T
53. cos 27 . Then y 72 22 45 3 5, and so sin 3 7 5 , tan 3 2 5 , csc T
7 7 5 , sec 7 ,
15 2
3 5
T
2 2 5.
and cot T
3 5 15
S T T T
54. tan 4. Then r 42 12 17, and so sin T4 , cos T1 , csc 417 , sec 17, and
17 17
cot 14 .
T T
55. If 2 3
3 , then sin 2 sin 3 2 and 2 sin 2 sin 3 3.
b c r s
56. If , then sin2 sin 2 3 and sin 2 sin 2 s 0890.
3 3 4 9
57. The area is 12 7 9 sin 72i s 300.

58. The area is 12 10 22 sin 10i s 191.
T
59. The area is 12 102 sin 60i 25 3 s 433.
T
60. The area is 12 132 sin 60i 1694 3 s 732.
r s
61. Let the angle be . Then A 12 x y sin " 16 12 5 7 sin % sin 32
35 % sin
1 32 1154, or
35
approximately 661i .
V
1 2 1 2 5 48 T T
62. Let the lengths of the equal sides be x. Then A 2 x sin % 24 2 x sin 6 % x 5
96 4 6 s 98 cm.
sin 6
4 . For the triangle defined by the two sides,
63. For the sector defined by the two sides, A1 12 r 2 12 22 120i 180i 3
T T
A2 12 ab sin 12 2 2 sin 120i 2 sin 60i 3. Thus the area of the region is A1 A2 43 3 s 246.
64. The area of the entire circle is r 2 122 144, the area of the sector is 12 r 2 12 122
3 24,
1 1 2 T
and the area of the triangle is 2 ab sin 2 12 sin 3 36 3, so the area of the shaded region is
T T
A1 A2 A3 144 24 36 3 120 36 3 s 4393.
h 1000 m
65. (a) tan , so h tan 1 . 6 km 1600 tan m.
1.6 kilometers 1 km
(b)
20i 60i 80i 85i
h 582 2771 9074 18,288
66. (a) Let the depth of the water be h. Then the cross-sectional area of the gutter is a trapezoid whose
height is h 10 sin . The bases are 10 and 10 2 10 cos 10 20 cos . Thus, the area is
b b2 10 10 20 cos
A 1 h 10 sin 100 sin 100 sin cos .
2 2
(b) (c)
200 130.0
100 129.5
0 129.0
0 1 1.00 1.05 1.10
From the graph, the largest area is achieved when

s 1047 rad s 60i .
SECTION 6.3 Trigonometric Functions of Angles 489
67. (a) From the figure in the text, we express depth and width in terms of . (b)
depth width
Since sin and cos , we have depth 20 sin 200
20 20
and width 20 cos . Thus, the cross-section area of the beam is
A depth width 20 cos 20 sin 400 cos sin .
0
(c) The beam with the largest cross-sectional area is the square beam, 0 1
T T
10 2 by 10 2 (about 1414 by 1414).
68. Using depth 20 sin and width 20 cos , from Exercise 65, we have strength k width depth2 . Thus
S k 20 cos 20 sin 2 8000k cos sin2 .
) 02 sin 2 3.6 2 sin
3 s 1145 m and the height is
69. (a) On Earth, the range is R
g 9.8
) 2 sin2 3.62 sin2 6 9 01653 m .
H 0
2g 2 9.8 16
3.62 sin 2 2
3 s 7 015 m and H 3.6 sin 6 s 1.013 m
(b) On the moon, R
16 2 16
U
600 T
70. Substituting, t 4 15 s 155 s.
5 sin 30 i
71. (a) W 302 038 cot 065 csc
10
(b) From the graph, it appears that W has its minimum value at about
0946 s 542i . 5
0
0 2
72. (a) We label the lengths L 1 and L 2 as shown in the (b)

n 14
3 2
.l
figure. Since sin and cos , we
L1 L2
7
have L 1 3 csc and L2 2 sec . Thus n
L L 1 L 2 3 csc 2 sec .
.| 0
(c) The minimum value of L is 7023. 0.0 0.5 1.0 1.5
n
(d) The minimum value of L is the shortest

distance the pipe must pass through.
73. We have sin k sin , where 594i and k 133. Substituting, sin 594i 133 sin
sin 594i
" sin s 06472. Using a calculator, we find that s sin1 06472 s 403i , so
133
4 2 s 4 403i 2 594i 424i .
74. sin 4i 00697564737 but sin 4 07568024953. Your partner has found sin 4i instead of sin 4 radians.
adj OP OP
75. cos O P . Since QS is tangent to the circle at R, -O R Q is a right triangle. Then
hyp OR 1
opp RQ hyp OQ
tan R Q and sec O Q . Since / SO Q is a right angle -S O Q is a right
adj OR adj OR
hyp OS adj SR
triangle and / O S R . Then csc O S and cot S R . Summarizing, we have
opp OR opp OR
sin P R , cos O P , tan R Q , sec O Q , csc O S , and cot S R .
r s 1 1
76. (a) sin2 cos2 1 % sin2 cos2 1 % tan2 1 sec2
cos2 cos2
r s 1 1
(b) sin2 cos2 1 % sin2 cos2 1 % 1 cot2 csc2
sin2 sin2
77. Let x represent the desired real number, in radians. We wish to find Z
Y
the smallest nonzero solution of the equation sin x sin 180x
. (Since
Zsin y
x is in radians, 180x
is in degrees.) Now if a is in degrees with
ZsinY
0 n a n 180 and sin a b, then either a b or 180 a b. The Y
graph shows that the first positive solution occurs when

180x 180x @
0 n 180, so we have 180 x.

180 180
Solving this equation, we get x s 3088. You can verify that the value of sin is the same whether
180 180
your calculator is set to radian mode or degree mode.
6.4 INVERSE TRIGONOMETRIC FUNCTIONS AND TRIANGLES

1. For a function to have an inverse, it must be one-to-one. To define the inverse sine function we restrict the domain of the
d e
sine function to the interval
2 2 .
d e
2. (a) The function sin1 has domain [1 1] and range 2 2 .
(b) The function cos1 has domain [1 1] and range [0 ].
b c
(c) The function tan1 has domain R and range 2 2 .
8 sin1 4
3. (a) sin1 10 6 cos1 3
(b) cos1 10 (c) tan1 86 tan1 43
5 5
r s
5 , we let cos1 5 and
4. To find sin cos1 13 13
complete the right triangle as shown. We find that
r s
5 12 .
sin cos1 13 13 n

T
5. (a) sin1 1
2 (b) cos1 0
2 (c) tan1 3
3
6. (a) sin1 0 0 (b) cos1 1 (c) tan1 0 0

r T s r T s
7. (a) sin1 22
4 (b) cos1 22 34 (c) tan1 1
4
r T s r s r T s
8. (a) sin1 23
3 (b) cos1 12 23 (c) tan1 3
3
r s
7. sin1 030 030469 10. cos1 02 s 177215 11. cos1 13 s 123096 12. sin1 65 s 098511
13. tan1 3 s 124905 14. tan1 4 s 132582 15. cos1 3 is undefined. 16. sin1 2 is undefined.
6 3 , so sin1 3 s 369i .
17. sin 10 7 , so tan1 7 s 213i .
18. tan 18
5 5 18
9 , so tan1 9 s 347i .
19. tan 13 20. sin 30 3 1 3 s 254i .
13 70 7 , so sin 7
21. sin 47 , so sin1 47 s 348i . 22. cos 89 , so cos1 89 s 273i .

SECTION 6.4 Inverse Trigonometric Functions and Triangles 491
d e
23. We use sin1 to find one solution in the interval 90i 90i . sin 23 " sin1 23 s 418i . Another solution with
between 0i and 180i is obtained by taking the supplement of the angle: 180i 418i 1382i . So the solutions of the
equation with between 0i and 180i are approximately 418i and 1382i .
d e
24. One solution is given by cos1 43 s 414i . This is the only solution, because cos x is one-to-one on 0 180i .
r s d e
25. One solution is given by cos1 25 s 1136i . This is the only solution, because cos x is one-to-one on 0 180i .
d e
26. tan1 20 s 871i , so the only solution in 0 180i is approximately 180 871 929i .
d e
27. tan1 5 s 787i . This is the only solution on 0 180i .
r s d e
28. One solution is sin1 45 s 531i . Another solution on 0 180i is approximately 180i 531i 1269i .
r s
29. To find cos sin1 45 , first let sin1 54 . Then is the number in the interval
d e
2 2 whose sine is 45 . We draw a right triangle with as one of its acute
angles, with opposite side 4 and hypotenuse 5. The remaining leg of the triangle is n
found by the Pythagorean Theorem to be 3. From the figure we get
r s
cos sin1 54 sin 35 .
r s r s
Another method: By the cancellation properties of inverse functions, sin sin1 54 is exactly 45 . To find cos sin1 45 , we
first write the cosine function in terms of the sine function. Let u sin1 54 . Since 0 n u n
2 , cos u is positive, and since
S U r s U r s2 T T
cos2 u sin2 u 1, we can write cos u 1 sin2 u 1 sin2 sin1 54 1 45 1 16 9
25 25 5 .
3
r s
Therefore, cos sin1 54 35 .
r s
30. To find cos tan1 54 , we draw a right triangle with angle , opposite side 4, and
r s

adjacent side 3. From the figure we see that cos tan1 34 cos 35 .
n

r s r s
12 , we draw a right triangle with
31. To find sec sin1 13 7 , we draw a right triangle with
32. To find csc cos1 25
angle , opposite side 12, and hypotenuse 13. From the angle , opposite side 7, and hypotenuse 25. From the
r s r s
12 sec 13 .
figure we see that sec sin1 13 figure we see that csc cos1 257 csc 25 .
5 24

n
n

r s r s
12 , we draw a right triangle with angle 34. To find cot sin1 2 , we draw a right triangle with angle
33. To find tan sin1 13 3
, opposite side 12, and hypotenuse 13. From the figure , opposite side 2, and hypotenuse 3. From the figure we
r s r s T
12 tan 12 .
we see that tan sin1 13 5 see that cot sin1 32 cot 25 .
n

n

r s
35. We want to find cos sin1 x . Let sin1 x, so sin x. We sketch a right

Y
triangle with an acute angle , opposite side x, and hypotenuse 1. By the
S n
Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have
r s Y"
S
cos sin1 x cos 1 x 2 .
Another method: Let u sin1 x. We need to find cos u in terms of x. To do so, we write cosine in terms of sine. Note
S d e
that
2 n u n 2 because u sin
1 x. Now cos u 1 sin2 u is positive because u lies in the interval .
2 2
S
Substituting u sin1 x and using the cancellation property sinsin1 x x gives cossin1 x 1 x 2 .
r s
36. We want to find sin tan1 x . Let tan1 x, so tan x. We sketch a right
Y"
Y
triangle with an acute angle , opposite side x, and adjacent side 1. By the
S n
Pythagorean Theorem, the remaining leg is x 2 1. From the figure we have
r s x
sin tan1 x sin S .
x2 1
r s
37. We want to find tan sin1 x . Let sin1 x, so sin x. We sketch a right

Y
triangle with an acute angle , opposite side x, and hypotenuse 1. By the
S n
Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have
r s Y"
x
tan sin1 x tan S .
1 x2
r s
38. We want to find cos tan1 x . Let tan1 x, so tan x. We sketch a right
Y"
Y
triangle with an acute angle , opposite side x, and adjacent side 1. By the
S n
Pythagorean Theorem, the remaining leg is x 2 1. From the figure we have
r s 1
1
cos tan x cos S .
x2 1
39. Let represent the angle of elevation of the ladder. Let h represent the height, in meters, that the ladder reaches on the
building. Then cos 1.8 1 03 s 1266 rad s 725i . By the Pythagorean Theorem, h 2 1.82 6 2 %
6 03 % cos
T T
h 36 3.24 32 .76 s 5.7 m.
40. Let be the angle of elevation of the sun. Then tan 29
36 0806 % tan
1 08 s 389i .
41. (a) Solving tan h2 for h, we have h 2 tan .

(b) Solving tan h2 for we have tan1 h2.
42. (a) Solving tan 15s. Solving for , we have tan1 15s.
(b) Set s 6 m to get tan 15 5
6 2 . Solving for , we have tan
1 5 s 682i .
2
SECTION 6.5 The Law of Sines 493
43. (a) Solving sin h204 for we have sin1 h204.

r s
(b) Set h 1 50 to get sin1 150 i
204 s 0826 rad s 473 .
6340 . Solving for , we get
44. (a) Since the radius of the earth is 634 0 k m, we have the relationship cos h6340
r s
cos1 h6 6340 .
340
(b) The arc length is s radius included angle 6340 2 1268 0.
r s
(c) s 1268 0 cos1 h6 6340
340
r s r s
(d) When h 16 0 we have s 12680 cos1 1606340 6340 1268 0 cos1 6340
6500 s 2819 2 km.
r s r s r s
(e) When s 392 0 we have 3920 12680 cos1 h6 3640 3920 1 6340 6340 3920
340 % 12680 cos h6340 % h6340 cos 12680 %
r s r s
3920 % h 63 4 0 sec 3920 6340 s 315 5 km.
h 6340 63 4 0 sec 12680 12680
t u t u
1 1 1 1
45. (a) sin sin s sin1 08102 s 541i
2 3 1 tan 10i 7 tan 10i
t u t u
1 1
(b) For n 2, sin1 s 483 i . For n 3, sin1 s 322i . For n 4,
5 tan 15i 7 tan 15i
t u
1 1 1
sin s 245i . n 0 and n 1 are outside of the domain for 15i , because s 3732
9 tan 15i tan 15i
1
and s 1244, neither of which is in the domain of sin1 .
3 tan 15i
1
46. Let sec1 x. Then sec x, as shown in the figure. Then cos , so
x
t u t u Y
1 1
cos1 . Thus, sec1 x cos1 , x o 1.
x x
n
In particular, sec1 2 cos1 21
3.
1
Let csc1 x. Then csc x, as shown in the figure. Then sin , so
x
t u t u Y
1 1
sin1 . Thus, csc1 x sin1 , x o 1.
x x
z
In particular, csc1 3 sin1 31 s 0340.
1
Let cot1 x. Then cot x, as shown in the figure. Then tan , so
x
t u t u
1 1
tan1 . Thus, cot1 x tan1 , x o 1.
x x
a
In particular, cot1 4 tan1 41 s 0245. Y
6.5 THE LAW OF SINES

sin A sin B sin C
1. In triangle ABC with sides a, b, and c the Law of Sines states that .
a b c
2. (a) The Law of Sines can be used to solve triangles in cases ASA or SSA.
(b) The Law of Sines can give ambiguous solutions in case SSA.
376 sin 57i
3. / C 180i 984i 246i 57i . x s 31875.
sin 984i
17 sin 1144i
4. / C 180i 375i 281i 1144i . x s 254.
sin 375i
267 sin 52i
5. / C 180i 52i 70i 58i . x s 248.
sin 58i
563 sin 67i
6. sin s 0646. Then s sin1 0646 s 403i .
802
36 sin 120i
7. sin C s 0693 % / C s sin1 0693 s 44i .
45
185 sin 50i
8. / C 180i 102i 28i 50i . x s 1449.
sin 102i
65 sin 46i 65 sin 20i
9. / C 180i 46i 20i 114i . Then a i s 51 and b s 24.
sin 114 sin 114i
2 sin 100i 2 sin 30i
10. / B 180i 30i 100i 50i . Then c i s 257 and a s 131.
sin 50 sin 50i
12 sin 44i
11. / B 68i , so / A 180i 68i 68i 44i and a s 899.
sin 68i
34 sin 80i 65 sin 69i
12. sin B s 0515, so / B sin1 0515 s 31i . Then / C 180i 80i 31i 69i and c s 62.
65 sin 80i
13. / C 180i 50i 68i 62i . Then 14. / C 180i 110i 23i 47i . Then
230 sin 50i 230 sin 68i 50 sin 23i 50 sin 110i
a i s 200 and b s 242. a s 267 and b s 642.
sin 62 sin 62i sin 47 i sin 47i
% %
c c
# $
c c
# $
15. / B 180i 30i 65i 85i . Then 16. / C 180i 95i 22i 63i . Then
a i s 50 and c s 9. b s 11169 and c s 9990.
sin 85 sin 85i sin 22i sin 22i
% %
c

c c c
# $ # $
17. / A 180i 51i 29i 100i . Then 18. / A 180i 100i 10i 70i . Then
a i s 89 and c s 71. a s 1097 and b s 203.
sin 29 sin 29i sin 100i sin 100i
% %
c c
c
# $
c
# $
15 sin 110i
19. Since / A 90i there is only one triangle. sin B s 0503 % / B s sin1 0503 s 30i . Then
28
/ C s 180i 110i 30i 40i , and so c
28 sin 40i / i / i
i s 19. Thus B s 30 , C s 40 , and c s 19.
sin 110
40 sin 37i
20. sin C s 0802 % C1 s sin 0822 s 534i or / C2 s 180i 534i s 1266i .
/ 1
30
30 sin 896i
If / C1 s 534i , then / B1 s 180i 37i 534i 896i and b1 s 498.
sin 37i
30 sin 164i
If / C2 s 1266i , then / B2 s 180i 37i 1266i 164i and b2 s 141.
sin 37i
Thus, one triangle has / B1 s 896i , / C1 s 534i , and b1 s 498; the other has / B2 s 164i , / C2 s 1266i , and
b2 s 141.
21. / A 125i is the largest angle, but since side a is not the longest side, there can be no such triangle.
45 sin 38i
22. sin B s 0660 % / B1 s sin1 0660 s 413i or / B2 s 180i 413i s 1387i .
42
42 sin 1007i
If / B1 s 413i , then / A1 s 180i 38i 413i 1007i and a1 s 67.
sin 38i
42 sin 33 i
If / B2 s 1387i , then / A2 s 180i 38i 1387i 33i and a2 s 39.
sin 38i
Thus, one triangle has / A1 s 1007 , / B1 s 413 , and a1 s 67; the other has / A2 s 33i , / B2 s 1387i , and a2 s 39.
i i
30 sin 25i
23. sin C s 0507 % / C1 s sin1 0507 s 3047i or / C2 s 180i 3947i 14953i .
25
25 sin 12453i
If / C1 3047i , then / A1 s 180i 25i 3047i 12453i and a1 s 4873.
sin 25i
25 sin 547 i
If / C2 14953i , then / A2 s 180i 25i 14953i 547i and a2 s 564.
sin 25i
Thus, one triangle has / A1 s 125i , / C1 s 30i , and a1 s 49; the other has / A2 s 5i , / C2 s 150i , and a2 s 56.
100 sin 30i
24. sin B 23 % / B1 s sin1 32 s 418i or / B2 s 180i 418i 1382i .
75
75 sin 1082i
If / B1 s 418i , then / C1 s 180i 30i 418i 1082i and c1 s s 1425.
sin 30i
75 sin 118i
If / B2 s 1382i , then / C2 s 180i 30i 1382i 118i and c2 s s 307.
sin 30i
Thus, one triangle has / B1 s 418i , / C1 s 1082i , and c1 s 1425; the other has / B2 s 1382i , / C2 s 118i , and
c2 s 307.
100 sin 50i
25. sin B s 1532. Since sin n 1 for all , there can be no such angle B, and thus no such triangle.
50
80 sin 135 i
26. sin B s 0566 % / B1 s sin1 0566 s 344i or / B2 s 180i 344i 1456i .
100
100 sin 106i
If / B1 s 344i , then / C s 180i 135i 344i 106i and c s s 259.
sin 135i
If / B2 s 180i 344i 1456i , then / A / B2 135i 1456i 180i , so there is no such triangle.
Thus, the only possible triangle is / B s 344i , / C s 106i , and c s 259.
26 sin 29i
27. sin A s 0840 % / A1 s sin1 0840 s 572i or / A2 s 180i 572i 1228i .
15
15 sin 938i
If / A1 s 572i , then / B1 180i 29i 572i 938i and b1 s s 309.
sin 29i
15 sin 281i
If / A2 s 1228i , then / B2 180 29i 1228i 282i and b2 s s 146.
sin 29i
Thus, one triangle has / A1 s 572i , / B1 s 938i , and b1 s 309; the other has / A2 s 1228i , / B2 s 282i , and
b2 s 146.
82 sin 58i
28. sin C s 0953, so / C1 s sin1 0953 s 724i or / C2 s 180 724i 1076i .
73
73 sin 496i
If / C1 s 724i then / A1 s 180i 58i 724i 496i and a1 s s 656.
sin 58i
73 sin 144i
If / C2 s 1076i , then / A2 s 180i 58i 1076i 144i and a2 s s 214.
sin 58i
Thus, one triangle has / A1 s 496 , / C1 s 724 , and a1 s 656; the other has / A2 s 144i , / C2 s 1076i , and
i i
a2 s 214.
sin 30i sin B 28 sin 30i
29. (a) From -ABC and the Law of Sines we get % sin B 07, so
20 28 20
/ B s sin1 07 s 44427i . Since -BC D is isosceles, / B / B DC s 44427i . Thus,
/ BC D 180i 2 / B s 91146i s 911i .
(b) From -ABC we get / BC A 180i / A / B s 180i 30i 44427i 105573i . Hence
/ DC A / BC A / BC D s 105573i 91146i 144i .
12 sin 25i
30. By symmetry, / DC B 25i , so / A 180i 25i 50i 105i . Then by the Law of Sines, AD s 525.
sin 105i
31. (a) Let a be the distance from satellite to the tracking station A in kilometers. Then the subtended angle at the satellite is
/ C 180i 93i 842i 28i , and so a
80 sin 842i
s 1629 km.
sin 28i
(b) Let d be the distance above the ground in kilometers. Then d 162 88 sin 87i s 1627 km.
x sin 48i sin 48i
32. (a) Let x be the distance from the plane to point A. Then i i i %
AB sin 180 32 48 sin 100i
sin 48i
x 8 s 6 03 km.
sin 100i
h
(b) Let h be the height of the plane. Then sin 32i " h 6 0 3 sin 32 i s 3 20 km.
x
AC AB AB sin 52i
33. / C 180i 82i 52i 46i , so by the Law of Sines, i i % AC , so substituting we
sin 52 sin 46 sin 46i
200 sin 52i
have AC s 219 m.
sin 46i
312 sin 486i
34. sin / ABC s 0444 % / ABC s sin1 0444 s 264i , and so / BC A s 180i 486i 264i 105i .
527
527 sin 105i
Then the distance between A and B is AB s 6785 m.
sin 486i
35. We draw a diagram. A is the position of the tourist and C is the top of the tower.
%
/ B 90i 56i 844i and so / C 180i 292i 844i 664i . Thus, by
c
105 sin 292i
the Law of Sines, the length of the tower is BC s 559 m.
sin 664i
c
$ #
36. % The situation is illustrated in the diagram.

AC 49. 5 sin 67i s 4 557 m , so using the Pythagorean Theorem
S
we can calculate B A 49.52 45572 s 19 3m and
S
c AD 54 2 4557 2 s 29 m. Thus the anchor points are
$ # & B A AD s 19 3 29 48.3 m apart.
37. The angle subtended by the top of the tree and the sun’s rays is / A 180i 90i 52i 38i . Thus the height of the tree
65 sin 30 i
is h s 53 m.
sin 38i
38. % Let x be the length of the wire, as shown in the figure. Since 12i , other angles
in ABC are 90i 58i 148i , and 180i 12i 148i 20i .
a x 100 sin 148i
Thus, % x 100 s 155 m.
sin 148i sin 20i sin 20i
Y
z $
`
#
c
39. Call the balloon’s position R. Then in -P Q R, we see that / P 62i 32i 30i , and / Q 180i 71i 32i 141i .
QR PQ sin 30i
Therefore, / R 180i 30i 141i 9i . So by the Law of Sines, i i % Q R 60 s 192 m.
sin 30 sin 9 sin 9i
40. $ Label the diagram as shown, and let the hill’s angle of elevation be . Then
sin sin 8i
z applying the Law of Sines to ABC, %
120 30
%
sin 4 sin 8i s 055669 " s sin1 055669 s 338i . But from -AB D,
/ B AD / B 8i 90i , so s 90i 8i 338i 482i .

c
`
# &
41. Let d be the distance from the earth to Venus, and let be the angle formed by sun, Venus, and earth. By the Law
sin sin 394i
of Sines, s 0878, so either s sin1 0878 s 614i or s 180i sin1 0878 s 1186i .
1 0723
d 0723
In the first case, % d s 1119 AU; in the second case,
sin 180i 394i 614i sin 394i
d 0723
% d s 0427 AU.
sin 180i 394i 1186i sin 394i
sin B sin 60i b sin 60i
42. (a) Applying the Law of Sines to -ABC, we get or sin B . Similarly, applying the Law of
b c c
sin B sin 120i r sin 120 i b r
Sines to -BC D gives or sin B . Since sin 120i sin 60i , we have %
r cd cd c cd
b c sin D sin 60i b sin 60i
(`). Similarly, from -ADC and the Law of Sines we have or sin D , and
r cd b d d
a sin 120i b sin 60i a sin 120i b d
from -B DC we have sin D . Thus, % . Combining this with
cd d cd a cd
b b c d cd b b ab r a
(`), we get 1. Solving for r, we find 1 % %
r a cd cd cd r a a b ab
ab
r .
ab
43
(b) r 12 cm.
43
(c) If a b, then r is infinite, and so the face is a flat disk.
43. By the area formula from Section 6.3, the area of -ABC is A 12 ab sin C. Because we are given a and the three
sin B sin A a sin B
angles, we need to find b in terms of these quantities. By the Law of Sines, %b . Thus,
b a sin A
t u
a sin B a 2 sin B sin C
A 12 ab sin C 12 a sin C .
sin A 2 sin A
1
ab sin C
Area of - ABC sin C
44. By the area formula from Section 6.3, 12 , because a and b are the same for both
Area of - A) B ) C ) ab sin C ) sin C )
2
triangles.
45. %
C B
% % %
$ $ C C C
B B{ B
# $ $ # $ #
a o b: One solution b a b sin A: Two solutions a b sin A: One solution a b sin A: No solution
/ A 30i , b 100, sin A 1 . If a o b 100 then there is one triangle. If 100 a 100 sin 30i 50, then there are
2
two possible triangles. If a 50, then there is one (right) triangle. And if a 50, then no triangle is possible.
6.6 THE LAW OF COSINES

1. For triangle ABC with sides a, b, and c the Law of Cosines states c2 a 2 b2 2ab cos C.
2. The Law of Cosines is required to solve triangles in cases SSS and SAS.
T
3. x 2 212 422 2 21 42 cos 39i 441 1764 1764 cos 39i s 834115 and so x s 834115 s 289.
T
T
T
37832 68012 42152
7. 37832 68012 42152 2 6801 4215 cos . Then cos s 0867 %
2 6801 4215
s cos1 0867 s 2989i .
15462 6012 12252
8. 15462 6012 12252 2 601 1225 cos . Then cos s 0359 %
2 601 1225
s cos1 0359 s 111i .
T
202 102 122 156
10. 202 102 122 2 10 12 cos . Then cos 065 % s cos1 065 s 13054i .
2 10 12 240
T
11. c2 102 182 2 10 18 cos 120i 100 324 360 cos 120i 604 and so c s 604 s 24576. Then
18 sin 120i
sin A s s 0634295 % / A s sin1 0634295 s 394i , and / B s 180i 120i 394i 206i .
24576
122 402 442
12. 122 402 442 2 40 44 cos B % cos B s 0964 % / B s cos1 0964 s 15i . Then
2 40 44
40 sin 155i
sin A s 0891 % / A s sin1 0891 s 63i , and so / C s 180i 15i 63i 102i .
12
SECTION 6.6 The Law of Cosines 499
T 4 sin 53i
13. c2 32 42 2 3 4 cos 53i 9 16 24 cos 53i s 10556 % c s 10556 s 32. Then sin B s 0983
325
% / B s sin1 0983 s 79i and / A s 180i 53i 79i 48i .
14. a 2 602 302 2 60 30 cos 70i 3600 900 3600 cos 70i s 326873
T 30 sin 70i
% a 326873 s 572. Then sin C s s 0493 %
572
/ C s sin1 0493 s 295i , and / B s 180i 70i 295i 805i .
2 252 222
15. 202 252 222 2 25 22 cos A % cos A 20 22522 s 0644 % / A s cos1 0644 s 50i . Then
i
sin B s 25 sin20499 s 0956 % / B s sin1 0956 s 73i , and so / C s 180i 50i 73i 57i .
2 122 162
16. 102 122 162 2 12 16 cos A % cos A 10 21216 078125 %
i
/ A s cos1 078125 s 386i . Then sin B s 12 sin 386 s 0749 %
10
/ B s sin1 0749 s 485i , and so / C s 180i 386i 485i 929i .
sin 40i s 0833 % / C s sin1 0833 s 564i or / C s 180i 564i s 1236i .
17. sin C 162125 1 2
sin 836i s 193.
If / C1 s 564i , then / A1 s 180i 40i 564i 836i and a1 125sin 40i
/ i / i i i i
If C2 s 1236 , then A2 s 180 40 1236 164 and a2 sin 125 sin 164i s 549.
40i
Thus, one triangle has / A s 836i , / C s 564i , and a s 193; the other has / A s 164i , / C s 1236i , and a s 549.
18. sin A 65 sin 52i s 1024. Since sin n 1 for all , there is no such / A, and hence there is no such triangle.
50
19. sin B 65 sin 55i s 1065. Since sin n 1 for all , there is no such / B, and hence there is no such triangle.
50
20. / A 180i 61i 83i 36i . Then b 735 sin 61i s 1094 and c 735 sin 83i s 1241.
sin 36i sin 36i
sin 35i s 2.
21. / B 180i 35i 85i 60i . Then x 3sin 60i
T
22. x 2 10i 18i 2 10 18 cos 40i 100 324 360 cos 40i s 148224 and so x s 148224 s 122.
i
23. x 50 sin 30
sin 100i s 254
2 102 112
24. 42 102 112 2 10 11 cos . Then cos 4 21011 205
220 s 0932 % s cos
1 0932 s 213i .
25. b2 1102 1382 2 110 138 cos 38i 12,100 19,044 30,360 cos 38i s 72200 and so b s 850. Therefore,
2 852 1282
using the Law of Cosines again, we have cos 1102110138 % s 8915i .
26. sin 10 sin 40i s 0803 % s sin1 0803 s 535i or s 180i 535i s 1265i , but 535i doesn’t fit the
8
picture, so s 1265i .
27. x 2 382 482 2 38 48 cos 30i 1444 2304 3648 cos 30i s 588739 and so x s 243.
28. / A 180i 98i 25i 57i . Then x 1000 sin 98i s 11808.
sin 57i
29. The semiperimeter is s 912152 18, so by Heron’s Formula the area is
T T
A 18 18 9 18 12 18 15 2916 54.
30. The semiperimeter is s 122
2 52 , so by Heron’s Formula the area is
U r sr sr s T T
A 52 52 1 52 2 52 2 15 15
16 4 s 0968.
31. The semiperimeter is s 789 2 12, so by Heron’s Formula the area is

T T T
A 12 12 7 12 8 12 9 720 12 5 s 268.
32. The semiperimeter is s 11100101
2 106, so by Heron’s Formula the area is
T T T
A 106 106 11 106 100 106 101 302,100 10 3021 s 550.
33. The semiperimeter is s 346 2 13

2 , so by Heron’s Formula the area is
U r sr sr s T T
A 13 13 3 13 4 13 6 455 455 s 533.
2 2 2 2 16 4
34. Both of the smaller triangles have the same area. The semiperimeter of each is s 255
2 6, so by Heron’s Formula the
T T T T
area of each is A 6 6 2 6 5 6 5 24 2 6, so the shaded area is 4 6 s 980.
35. We draw a diagonal connecting the vertices adjacent to the 100i angle. This forms two triangles. Consider the triangle with
sides of length 5 and 6 containing the 100i angle. The area of this triangle is A1 12 5 6 sin 100i s 1477. To use
Heron’s Formula to find the area of the second triangle, we need to find the length of the diagonal using the Law of Cosines:
c2 a 2 b2 2ab cos C 52 62 2 5 6 cos 100i s 71419 " c s 845. Thus the second triangle has semiperimeter
8 7 845 T
s s 117255 and area A2 117255 117255 8 117255 7 117255 845 s 2600. The area
2
of the quadrilateral is the sum of the areas of the two triangles: A A1 A2 s 1477 2600 4077.
36. We draw a line segment with length x bisecting the 60i angle to create two triangles. By the Law of Cosines,
T T T
32 42 x 2 2 4x cos 30i % x 2 4 3x 7 0. Using the Quadratic Formula, we find x 2 3 5. The minus
sign provides the correct length of about 123 (the other solution is about 57, which corresponds to a convex quadrilateral
T T
342 3 5
with the same side lengths), so the semiperimeter of each triangle is s and the total area of the figure
2
U r T T s
is A 2 s s 3 s 4 s 2 3 5 s 246.
37. Label the centers of the circles A, B, and C, as in the figure. By the Law of
AB 2 AC 2 BC 2 92 102 112
% $ Cosines, cos A 13 " / A s 7053i .
2 AB AC 2 9 10
sin 7053i sin B sin C
Now, by the Law of Sines, . So
11 AC AB

# sin B 10 i
11 sin 7053 s 085710 " B s sin
1 085710 s 5899i and
9 sin 7053i s 077139 " C s sin1 077139 s 5048i . The area of

sin C 11
ABC is 12 AB AC sin A 12 9 10 sin 7053i s 42426.

i
The area of sector A is given by S A R 2 42 7053 s 9848. Similarly, the areas of sectors B and C
360i 360i
are S B s 12870 and SC s 15859. Thus, the area enclosed between the circles is A ABC S A S B SC "
A s 42426 9848 12870 15859 s 385 cm2 .
T
38. By the Law of Cosines we have a 2 62 42 2 6 4 cos 45i 52 24 2,
T rT T s
b2 62 x 2 2 6 x cos 30i x 2 6 3x36, and c2 x 2 42 2 x 4 cos 30i 45i x 2 2 2 6 x16.
T r T s rT T s
By the Pythagorean Theorem, a 2 b2 c2 , so we have 52 24 2 x 2 6 3x 36 x 2 2 2 6 x 16
r T s
T T rT T s T r T T T s 12 3 2
% 26 12 2 3 3x 18 2 6 x 8 % 36 12 2 3 3 2 6 x, so x T T T .
3 3 2 6
39. Let c be the distance across the lake, in kilometers. Then c2 2822 3562 2 282 356 cos 403i s 5313 %
c s 230 km.
SECTION 6.6 The Law of Cosines 501
& %
40. Suppose ABC D is a parallelogram with AB DC 5, AD BC 3,
and / A 50i (see the figure). Since opposite angles are equal in a

parallelogram, it follows that / C 50i , and cM
/ B / D 360i 100i 260i . Thus, / B / D

260i # $
130i .
2
T
By the Law of Cosines, AC 2 32 52 2 3 5 cos 130i " AC 9 25 30 cos 130i s 73. Similarly,
S
B D 32 52 2 3 5 cos 50i s 38.
41. In half an hour, the faster car travels 25 kilometers while the slower car travels 15 kilometers. The distance
between them is given by the Law of Cosines: d 2 252 152 2 25 15 cos 65i "
S T
d 252 152 2 25 15 cos 65i 5 25 9 30 cos 65i s 231 km.
42. Let x be the car’s distance from its original position. Since the car travels
Y
at a constant speed of 40 kilometers per hour, it must have traveled 40 kilometers
east, and then 20 kilometers northeast (which is 45i east of “due north”). From z c
the diagram, we see that / 135i , so
S T
x 202 402 2 20 40 cos 135i 10 4 16 16 cos 135i s 560 k m.
43. The pilot travels a distance of 1000 15 1500 kilometers in her original c

direction and 1000 2 200 0 kilometer s in the new direction. Since
she makes a course correction of 10 i to the right, the included angle is E
180i 10i 170i . From the figure, we use the Law of Cosines to get
the expression d 2 1500 2 2000 2 2 1500 2000 cos 170i s 12,158,846 52, so d s 3487 kilometers. Thus,
the pilot’s distance from her original position is approximately 348 7 k m.
44. Let d be the distance between the two boats in kilometers. After one hour, the boats have traveled distances of
30 km and 26 km. Also, the angle subtended by their directions is 180i 50i 70i 60i . Then
T
d 2 302 262 2 30 26 cos 60i 796 % d s 796 s 282. Thus the distance between the two boats is about
28 kilometers.
45. (a) The angle subtended at Egg Island is 100i . Thus using the Law of Forrest Island
Cosines, the distance from Forrest Island to the fisherman’s home port is
x 2 302 502 2 30 50 cos 100i

900 2500 3000 cos 100i s 3920945 c

T Y
and so x s 3920945 s 6262 kilometers.
(b) Let be the angle shown in the figure. Using the Law of Sines, c
50 sin 100i Egg Island c
a
sin s 07863 % s sin1 07863 s 518i . Then n
6262 c
90i 20i 518i 182i . Thus the bearing to his home port is
Home Port
S 182i E.
46. (a) In 30 minutes the pilot flies 160 kilometers due east, so using the Law of $
Cosines we have
x 2 1602 4802 2 160 480 cos 40 i
Y
1602 1 9 6 cos 40i s 1602 5404.
S c
Thus, x s 1602 5404 s 371 9, and so the pilot is 371 9 kilometers from c n
#
his destination.
480 sin 40 i
(b) Using the Law of Sines, sin s 0829 %
3719
s sin1 0829 s 56i . However, since 90i , the angle we seek is
180i 56i 124i . Hence the bearing is 124i 90i 34i , that is,
N 34i E.
47. The largest angle is the one opposite the longest side; call this angle . Then by the Law of Cosines,
362 222 442
442 362 222 2 36 22 cos % cos 009848 " s cos1 009848 s 96i .
2 36 22
48. Let be the angle formed by the cables. The two tugboats and the barge form a triangle: the side opposite has a length of
36 m and the other two sides have lengths of 64 and 69 m. Therefore, 36 2 64 2 69 2 2 64 69 cos %
64 2 69 2 36 2
cos " cos 0856 1 " s cos1 08561s 31 i.
2 64 69
49. Let d be the distance between the kites. Then d 2 s 1142 1262 2 114 126 cos 30i "
S
d s 1142 1262 2 114 126 cos 30i s 63 m.
50. Let x be the length of the wire and the angle opposite x, as shown in the figure.
Since the mountain is inclined 32i , we must have
180i 90i 32i 122i . Thus, Y
S
x 172 17 2 2 17 38 cos 122i s 49 m. n

c
1020
51. Solution 1: From the figure, we see that 106i and sin 74i %
b
1020
b s 1061. Thus, x2 2402 10612 2 240 1061 cos 106i "
sin 74i
S Y
x 2402 10612 2 240 1061 cos 106i " x s 1151 m . C
1020 1020
Solution 2: Notice that tan 74i % a s 292 5. By the
a tan 74i
a c
Pythagorean Theorem, x 2 a 2 4 02 10202 . So
T
x 292 5 2402 10202 s 1151 m.

52. Let the woman be at point A, the first landmark (at 62i ) be at point B, and the other landmark be at point C. We
345 345 345
want to find the length BC. Now, cos 62i % AB s 735. Similarly, cos 54 i %
AB cos 62i AC
345
AC s 587. Therefore, by the Law of Cosines, BC 2 AB 2 AC 2 2 AB AC cos 43i "
cos 54i
S
BC 735 2 587 2 2 735 587 cos 43i " BC s 504. Thus, the two landmarks are roughly 504 meters apart.
CHAPTER 6 Review 503
T abc 34 44 57
53. By Heron’s formula, A s s a s b s c, where s 67.5. Thus,
2 2
T
A 67.5 67.5 34 6 7 . 5 44 67.5 57 s 747 m 2 . Since the land value is $20 0 per square meter , the value of the
lot is approximately 74 7 20 0 $149,400.
sin B sin 465i
54. Having found a s 132 using the Law of Cosines, we use the Law of Sines to find / B: %
105 132
105 sin 465i
sin B s 0577. Now there are two angles / B between 0i and 180i which have sin B 0577, namely
132
/ B s 352i and / B s 1448i . But we must choose / B , since otherwise / A / B 180i .
1 2 1
sin C sin 465i 180 sin 465i
Using the Law of Sines again, % sin C s 0989, so either / C s 815i
180 132 132
or / C s 985i . In this case we must choose / C s 985i so that the sum of the angles in the triangle is
/ A / B / C s 465i 352i 985i s 180i . (The fact that the angles do not sum to exactly 180i , and the
discrepancies between these results and those of Example 3, are due to roundoff error.)
The method in this exercise is slightly easier computationally, but the method in Example 3 is more reliable.
55. In any ABC, the Law of Cosines gives a 2 b2 c2 2bccos A, b2 a 2 c2 2accos B, and c2 a 2 b2 2abcos C.
Adding the second and third equations gives
b2 a 2 c2 2ac cos B
c2 a 2 b2 2ab cos C
b2 c2 2a 2 b2 c2 2a c cos B b cos C
Thus 2a 2 2a c cos B b cos C 0, and so 2a a c cos B b cos C 0.
Since a / 0 we must have a c cos B b cos C 0 % a b cos C c cos B . The other laws follow from the symmetry
of a, b, and c.
CHAPTER 6 REVIEW
s 052 rad
1. (a) 30i 30 180 7 s 183 rad
2. (a) 105i 105 180
6 12
5 s 262 rad
(b) 150i 150 180 2 s 126 rad
(b) 72i 72 180
6 5
s 035 rad
(c) 20i 20 180 9 s 707 rad
(c) 405i 405 180
9 4
5 s 393 rad
(d) 225i 225 180 7 s 550 rad
(d) 315i 315 180
4 4
3. (a) 56 rad 56 180

150
i 4. (a) 53 rad 53 180
300
i
(b) 180
9 rad 9 20
i (b) 109 rad 109 180
200
i
(c) 43 rad 43 180

240
i (c) 5 rad 5 180 900
s 2865
i
(d) 4 rad 4 180 720

s 2292
i (d) 113 rad 113 180
660
i
5. r 10 m, 25 rad. Then s r 10 25 4 s 126 m.

s 7 28 rad s 1604i
6. s 7 cm, r 25 cm. Then 25
r
5 rad. Then r s 25 18 90 s 286 m.
7. s 25 m , 50 i 50 180 18 5
13 rad. Then r s 13 18 18 m.
8. s 13 m, 130i 130 180 18 13
9. Since the diameter is 70 cm , r 35cm . In one revolution, the arc length (distance traveled) is s r 2 35 70 cm.
The total distance traveled is 9 6 k m/h 05 h 48 k m 48 k m 10 0 0 m /k m 100 cm /m 4 ,8 00,0 00. The number of
1 rev
revolution is 4 ,8 00,0 00 in s 21,827 rev. Therefore the car wheel will make approximately 21,827 revolutions.
70 cm
s 3920 s 35448i and so the angle
10. r 6340 kilometers, s 3920 kilometers. Then s 0619 rad 0619 180
r 6340
is approximately 35i 4 .
11. r 5 m, 2 rad. Then A 12 r 2 12 52 2 25 m2 .

b c s 18,151 m2
12. A 12 r 2 12 2002 52i 180i
2A 2 125
13. A 125 m 2 , r 25 m . Then 2 250
625 04 rad s 229
i
r 252
U V
2A 100
14. A 50 m2 and 116 rad. Thus, r 11 s 42 m.

6
150 2 rad
15. The angular speed is $ 300 rad/min s 9425 rad/min. The linear speed is
1 min
150 2 2 0
) 6000 cm /min s 18,849 6 cm /min s 188 5 m /min.
1
16. (a) The angular speed of the engine is $e 35002 rad 7000 rad/min s 21,9911 rad/min.
1 min
$e
(b) To find the angular speed $* of the wheels, we calculate g $ *
7000$rad/min
*
09 %
$* s 77778 rad/min s 24,4346 rad/min.
(c) The speed of the car is the angular speed of the wheels times their radius:
77778 rad 28 cm 60 min 1 k m
min 1 hr 100,000 cm s 410 5 k m/h.
S T T T
17. r 52 72 74. Then sin T5 , cos T7 , tan 57 , csc 574 , sec 774 , and cot 75 .
74 74
S T T T
3 , cos 91 , tan T3 , csc 10 , sec T10 , and cot 91 .
18. x 102 32 91. Then sin 10 10 3 91 3 91
x
19. cos 40i % x 5 cos 40i s 383, and 5y sin 40i % y 5 sin 40i s 321.
5
2
20. cos 35i % x cos235i s 244, and tan 35i 2y % y 2 tan 35i s 140.
x
1
21. sin 20i % x sin120i s 292, and xy cos 20i % y cosx20i s 09397 2924 s 311.
x
x
22. cos 30i % x 4 cos 30i s 346, and sin 30i xy % y x sin 30i 346 05 s 173.
4
23. A 90i 20i 70i , a 3 cos 20i s 2819, and 24. C 90i 60i 30i , cos 60i 20a %
b 3 sin 20i s 1026. a 20 cos 60i 40, and tan 60i b20 %
% b 20 tan 60i s 3464.
C
# c %
B
c
c B
C
$
c
# $
S S
25. c 7 s 02960 s 170i ,
252 72 24, A sin1 24 26. b 5 s 03948 s 226i ,
122 52 13, A sin1 13
and C sin1 24 i
25 s 12870 s 737 . and C sin1 12 i
13 s 11760 s 674 .
% %

C

# D $
# $

1 1 1 1
27. tan %a cot , sin % b csc
a tan b sin
h
28. Let h be the height of the tower in meters. Then tan 2881i % h 1000 tan 2881i s 550 m.
1000
29. One side of the hexagon together with radial line segments through its endpoints
forms a triangle with two sides of length 8 m and subtended angle 60i . Let x be the
Y c length of one such side (in meters). By the Law of Cosines,

x 2 82 82 2 8 8 cos 60i 64 % x 8. Thus the perimeter of the
hexagon is 6x 6 8 48 m.
30. As the crankshaft moves in its circular pattern, point Q is Z Z
determined by the angle , namely it has coordinates

Q 5 cos , 5 sin . We split the triangle into two right triangles
-O Q R and -P Q R, as shown in the figure. Let h be the height of
the piston. We consider two cases, 0i n 180i and I

I
180i n 360i .
4 3 cos
n n
sin
If 0i n 180i , then h is the sum of O R and R P. Using the
n n
T
1 Y 1 Y
Pythagorean Theorem, we find R P 20 2 5 cos 2 , while
3 cos n s in n 4
O R is the y-coordinate of the point Q, 2 sin . Thus
S
h 400 25 cos 2 5 sin .
S
If 180i n 360i , then h is the difference between R P and R O. Again, R P 400 25 cos 2 and O R is the
S
y-coordinate of the point Q, 2 sin . Thus h 400 25 cos 2 5 sin . Since sin 0 for 180i 360i , this also
S
reduces to h 400 25 cos 2 5 sin .
S
Since we get the same result in both cases, the height of the piston in terms of is h 400 25 cos 2 5 sin .
r

31. Let r represent the radius, in kilo meters , of the moon. Then tan , 0518i % r r 379,040 tan 0 259i
2 r AB
379,040 tan 0259i
% r 1 tan 0259i 379,040 tan 0259i % r s 1722 and so the radius of the moon is
1 tan 0259i
roughly 1722 kilometers.
32. Let d1 represent the horizontal distance from a point directly below the plane to the closer ship in meters , and d2 represent the
10,500 10,500 10,500
horizontal distance to the other ship in meters . Then tan 52i % d1 , and similarly tan 40i %
d1 tan 52i d2
10,500 10,500 10,500
d2 . So the distance between the two ships is d2 d1 s 4320 m .
tan 40i tan 40i tan 52i
T T
33. sin 315i sin 45i T1 22 34. csc 94 csc
4 2
2
T
35. tan 135i tan 45i 1 36. cos 56 cos
6 2
3
r s T T
37. cot 223 cot 23 cot T1
3 3 3
3 38. sin 405i sin 45i T1 22
2
T
39. cos 585i cos 225i cos 45i T1 22 40. sec 223 sec 43 sec
3 2
2
T T
41. csc 83 csc 23 csc T2 2 3
3 3 3 42. sec 136 sec 2 3
6 3
T
43. cot 390i cot 30i cot 30i 3 44. tan 234 tan 34 tan
4 1
T T
45. r 52 122 169 13. Then sin 12 5 12 13 13
13 , cos 13 , tan 5 , csc 12 , sec 5 , and
5 .
cot 12
46. If is in standard position, then the terminal point of on the unit circle is simply cos sin . Since the terminal point is
r T s
given as 23 12 , sin 12 .
T T T T
47. y 3x 1 0 % y 3x 1, so the slope of the line is m 3. Then tan m 3 % 60i .
S T
48. 4y 2x 1 0 % y 12 x 14 . The slope of the line is m 12 . Then tan m 12 and r 12 22 5. So
T T
sin T1 , cos T2 , tan 12 , csc 5, sec 25 , and cot 2.
5 5
S
S sin 1 cos2
49. Since sin is positive in quadrant II, sin 1 cos2 and we have tan .
cos cos
1 1
50. sec S (because cos 0 in quadrant III).
cos 1 sin2
sin2 sin2
51. tan2
cos2 1 sin2
1 1 sin2 1
52. csc2 cos2 cos2 1
sin2 sin2
sin2
T T T T
53. tan 37 , sec 43 . Then cos 34 and sin tan cos 47 , csc T4 4 7 7 , and cot T3 3 7 7 .
7 7
9
sin 41
54. sec 41 41 9 40 9
40 , csc 9 . Then sin 41 , cos 41 , tan cos 40 40 , and cot 9 .
40
41
S T
55. sin 35 . Since cos 0, is in quadrant II. Thus, x 52 32 16 4 and so cos 45 , tan 34 ,
csc 53 , sec 54 , cot 43 .
56. sec 13 5 and tan 0. Then cos 13 5 , and must be in quadrant III " sin 0. Therefore,
S T
sin 1 cos2 1 169 25 12 , csc 13 , tan sin 12 , and cot 5 .
13 12 cos 5 12
4 T
57. tan 12 . sec2 1 tan2 1 14 54 % cos2 " cos 45 T2 since
5 5
sin r s
cos 0 in quadrant II. But tan 1 1
2 % sin 2 cos 2 T1 2 1
T . Therefore,
cos 5 5
r s T
1 2
sin cos T T T 5 . 1 5
5 5 5
1 1 T
sin 1 sin 1 sin 1 2 3 T
58. sin 12 for in quadrant I. Then tan sec S T2 3.
cos cos cos 1 sin2 1 1 2
2
59. By the Pythagorean Theorem, sin2 cos2 1 for any angle .
T T
60. cos 23 and 5 10 5 5 3
2 . Then 6 " 2 6 3 . So sin 2 sin 3 sin 3 2 .
T T
61. sin1 23
3 62. tan1 33
6
63. Let u sin1 52 and so sin u 25 . Then from the triangle, 64. Let u cos1 83 then cos u 38 . From the triangle, we
r s r s T
tan sin1 52 tan u T2 . have sin cos1 83 sin u 855 .
21
V

V

65. Let tan1 x % tan x. Then from the 66. Let sin1 x. Then sin x. From the triangle, we
r s x r s
triangle, we have sin tan1 x sin S . 1
have sec sin1 x sec S .
1 x2 1 x2
Y" Y Y
n
n

Y"
x rx s x rx s
67. cos " cos1 " tan1
68. tan
3 3 2 2
10 sin 30i
i i i i
69. / B 180 30 80 70 , and so by the Law of Sines, x s 532.
sin 70i
2 sin 45i
70. x s 146
sin 105i
71. x 2 1002 2102 2 100 210 cos 40i s 21,926133 % x s 14807
20 sin 60i
72. sin B s 0247 % / B s sin1 0247 s 1433i . Then / C s 180i 60i 1433i 10567i , and so
70
70 sin 10567i
xs s 7782.
sin 60i
T
73. x 2 22 82 2 2 8 cos 120i 84 % x s 84 s 917
4 sin 110i 6 sin 3121i
74. sin B s 0626 % / B s 3879i . Then / C s 180i 110i 3879i 3121i , and so x s s 33.
6 sin 110i
t u
sin sin 25i 23 sin 25i 23 sin 25i
75. By the Law of Sines, " sin " sin1 s 541i or
23 12 12 12
s 180i 541i 1259i .
t u
sin sin 80i 4 sin 80i 4 sin 80i
76. By the Law of Sines, " sin " sin1 s 520i .
4 5 5 5
1202 1002 852
77. By the Law of Cosines, 1202 1002 852 2 100 85 cos , so cos s 016618. Thus,
2 100 85
s cos1 016618 s 804i .
sin A sin 10i 5 sin 10i
78. We first use the Law of Sines to find / A: " sin A s 02894, so A s sin1 02894 s 168i
5 3 3
or A s 180i 168i 1632i . Therefore, s 180i 10i 168i 1532i or 180i 10i 1632i 68i .
79. After 2 hours the ships have traveled distances d1 40 mi and d2 56 km. The subtended angle is
180i 32i 42i 106i . Let d be the distance between the two ships in kilo meter s. Then by the Law of Cosines,
d 2 402 562 2 40 56 cos 106i s 5970855 % d s 773 kilo meter s.
80. Let h represent the height of the building in meters , and x the horizontal distance from the building to point B. Then
h h h
tan 241i and tan 302i % x h cot 302i . Substituting for x gives tan 241i %
x 18 0 x h cot 302i 18 0
18 0 tan 241 i
h tan 241i h cot 302i 18 0 % h s 348 m .
1 tan 24 1i cot 30 2i
81. Let d be the distance, in kilometers, between the points A and B . Then by the Law of Cosines,
d 2 322 562 2 32 56 cos 42i s 14966 % d s 39 k m.
120 sin 689i
82. / C 180i 423i 689i 688i . Then b s 12008 kilo meters . Let d be the shortest distance,
sin 688i
in kilometers, to the shore. Then d b sin A s 12008 sin 423i s 808 kilometers .
83. A 12 ab sin 12 8 14 sin 35i s 3212
T abc 568
84. By Heron’s Formula, A s s a s b s c, where s 95. Thus,
2 2
T
A 95 95 5 95 6 95 8 s 1498.
CHAPTER 6 TEST
11 rad. 135i 135 3 rad.
1. 330i 330 180 6 180 4
2. 43 rad 43 180 i 180 234

240 . 13 rad 13 s 745
i
T
3. (a) The angular speed is 4. (a) sin 405i sin 45i T1 22
2
120 2 rad T
$ 240 rad/min s 75398 rad/min. (b) tan 150i tan 30i T1 33
1 min 3
(b) The linear speed is (c) sec 53 sec
3 2
120 2 5
) 1200 m /min (d) csc 52 csc
2 1
1
s 3769 9 m /min
s 226 k m/h
rT s
2 13 3 T
S T 2 2 26 6 13
5. r 32 22 13. Then tan sin T T .
3 13 3 13 39
a b
6. sin % a 24 sin . Also, cos % b 24 cos .
24 24
S T
7. cos 13 and is in quadrant III, so r 3, x 1, and y 32 12 2 2. Then
1 T T
tan cot csc tan csc 1 T3 2 T 23 43 2.
tan 2 2 2 2 4
5 , tan 5 . Then sec 1 1 sin 1 5 13 13

8. sin 13 12 tan .
cos cos sin sin 12 5 12
S S
9. sec2 1 tan2 % tan sec2 1. Thus, tan sec2 1 since tan 0 in quadrant II.
h
10. tan 73i " h 1.8 tan 73i s 5 9 m .
1.8
x rx s
11. (a) tan " tan1
4 4
t u
3 3
(b) cos " cos1
x x
Surveying 509
S
9 so tan u 9 . From the triangle, r
12. Let u tan1 40 92 402 41. So
40
r s
1 9 40 V
cos tan 40 cos u 41 .

13. By the Law of Cosines, x 2 102 122 2 10 12 cos 48i s 8409 % x s 91.
230 sin 69i
14. / C 180i 52i 69i 59i . Then by the Law of Sines, x s 2505.
sin 59i
h x h
15. Let h be the height of the shorter altitude. Then tan 20i % h 50 tan 20i and tan 28i % x h 50 tan 28i
50 50
% x 50 tan 28i h 50 tan 28i 50 tan 20i s 84.
15 sin 108i
16. Let / A and / X be the other angles in the triangle. Then sin A s 0509 % / A s 3063i . Then
28
/ X s 180i 108i 3063i s 4137i , and so x s
28 sin 4137i
s 195.
sin 108i
82 62 92
17. By the Law of Cosines, 92 82 62 2 8 6 cos " cos s 01979, so s cos1 01979 s 786i .
2 8 6
18. We find the length of the third side x using the Law of Cosines: x 2 52 72 2 5 7 cos 75i s 5588 " x s 7475.
sin sin 75i 5 sin 75i
Therefore, by the Law of Sines, s " sin s s 06461, so s sin1 06461 s 402i .
5 7475 7475
50 72 . A triangle 1 r r sin 1 102 sin 72i . Thus, the area of the
19. (a) A sector 12 r 2 12 102 72 180 180 2
r s 2
shaded region is A shaded A sector A triangle 50 72 i s 153 m2 .
180 sin 72
(b) The shaded region is bounded by two pieces: one piece is part of the triangle, the other is part of the circle.
S T
The first part has length l 102 102 2 10 10 cos 72i 10 2 2 cos 72i . The second has length
4. Thus, the perimeter of the shaded region is p l s 10T2 2 cos 72i 4 s 243 m.
s 10 72 180
92 132 202
20. (a) If is the angle opposite the longest side, then by the Law of Cosines cos 06410. Therefore,
2 9 20
cos1 06410 s 1299i .
(b) From part (a), s 1299i , so the area of the triangle is A 12 9 13 sin 1299i s 449 units2 . Another way to find
T abc 9 13 20
the area is to use Heron’s Formula: A s s a s b s c, where s 21. Thus,
2 2
T T
A 21 21 20 21 13 21 9 2016 s 449 units2 .
21. Label the figure as shown. Now / 85i 75i 10i , so by the Law of Sines,
x 30 sin 75i h z
% x 30 . Now sin 85i %
sin 75i sin 10i sin 10i x Y I
sin 75i
h x sin 85i 3 0 sin 85i s 166 .
sin 10i c c

FOCUS ON MODELING Surveying

1. Let x be the distance between the church and City Hall. To apply the Law of Sines to the triangle with vertices at City Hall,
the church, and the first bridge, we first need the measure of the angle at the first bridge, which is 180i 25i 30i 125i .
x 086 086 sin 125i
Then % x s 14089. So the distance between the church and City Hall is about
141 kilometers.
510 Focus on Modeling
2. To find the distance z between the fire hall and the school, we use the bank
distance found in the text between the bank and the cliff. To find z we
first need to find the length of the edges labeled x and y. c
In the bank-second bridge-cliff triangle, the third angle is

155 sin 50i
180i 50i 60i 70i , so x i s 126. c second bridge
sin 70 Y
c
In the second bridge-school-cliff triangle, the third angle is c
cliff c
126 sin 45i
180i 80i 55i 45i , so y s 109. c
sin 55i Z
Finally, in the school-fire hall-cliff triangle, the third angle is c

c c
fire hall
109 sin 80i [ school
180i 45i 80i 55i , so z s 131.
sin 55i
Thus, the fire hall and the school are about 131 kilometers apart.
3. First notice that / D BC 180i 20i 95i 65i and / D AC 180i 60i 45i 75i .
AC 20 20 sin 45i
From -AC D we get i i % AC s 146i . From -BC D we get
sin 45 sin 75 sin 75i
BC 20 20 sin 95i
% BC s 220. By applying the Law of Cosines to -ABC we get
T
AB 2 AC 2 BC 2 2 AC BC cos 40i s 1462 2202 2146220cos 40i s 205, so AB s 205 s 143 m.
Therefore, the distance between A and B is approximately 143 m.
4. Let h represent the height in meters of the cliff, and d the horizontal distance to the cliff. The third horizontal
200 sin 516i h
angle is 180i 694i 516i 59i and so d i s 182857. Then tan 331i %
sin 59 d
i i
h d tan 331 s 182857 tan 331 s 1192 m.
BC AB
5. (a) In -ABC, / B 180i , so / C 180i 180i . By the Law of Sines,
sin sin
sin d sin
" BC AB .
sin sin
d sin h h
(b) From part (a) we know that BC . But sin % BC . Therefore,
sin BC sin
d sin h d sin sin
BC "h .
sin sin sin
d sin sin 800 sin 25i sin 29i
(c) h s 2350 m
sin sin 4i
6. Let the surveyor be at point A, the first landmark (with angle of depression 42i ) be at point B, and the other landmark
2430 2430 2430
be at point C. We want to find BC . Now sin 42i % AB s 36316. Similarly, sin 39i
AB sin 42i AC
2430
% AC s 38613. Therefore, by the Law of Cosines, BC 2 AB 2 AC 2 2 AB AC cos 68i "
sin 39i
S
BC 363162 386132 2 36316 38613 cos 68i s 4194. Thus, the two landmarks are approximately 4194 m
apart.
Surveying 511
7. We start by labeling the edges and calculating the remaining angles, as shown in the first figure. Using the Law of Sines,
a 150 150 sin 29i b 150 150 sin 91i
we find the following: % a s 8397, % b s 17318,
sin 29i sin 60i sin 60i sin 91i sin 60i sin 60i
c 17318 17318 sin 32 i d 17318 17318 sin 61 i
%c s 9190, %e s 15167,
e 15167 15167 sin 41i f 15167 15167 sin 88 i
%e s 12804, % f s 19504,
g 19504 19504 sin 50i h 19504 19504 sin 38i
i i %g i s 14950, and i i %h s 12015. Note that
sin 50 sin 92 sin 92 sin 38 sin 92 sin 92i
we used two decimal places throughout our calculations. Our results are shown (to one decimal place) in the second figure.
D F
c c c c
c c
E
C G H

c c
c c c c
B I
8. Answers will vary. Measurements from the Great Trigonometric Survey were used to calculate the height of Mount Everest
to be exactly 8839 m , but in order to make it clear that the figure was considered accurate to within a meter , the height was
published as 8839.8 m . The accepted figure today is 88 48 m .
Another random document with
no related content on Scribd:
Sketches of the Manners, Customs, and History
of the Indians of America.
CHAPTER I.
First discoveries of Columbus.—The first interview between the
Spaniards and the Indians.—Simplicity of the Indians.—Their
appearance and manners.—Cuba discovered.—
Disappointment of Columbus in his search for gold.—Sails for
Hayti.
It was on the 12th of October, 1492, that Columbus first set his
foot on the shores of the New World. He landed at a small island
belonging to the Bahamas, which he named San Salvador. With a
drawn sword in his hand, he took possession of the country for his
sovereigns, Ferdinand and Isabella of Spain. I always regretted that
Columbus unsheathed the sword. He only intended it as a ceremony,
but it has proved a fatal reality to the poor Indians. The sword has
almost always been unsheathed between them and their christian
invaders.
It is my purpose, in the course of my story, to give a brief view of
the past and present condition of the Red Men of this western world.
I shall first notice the people of the West India Islands; then of South
America; then of North America; giving such sketches and
descriptions as can be relied upon for truth, and which combine
entertainment with instruction.
Irving, in his history of Columbus, thus beautifully narrates the first
interview between the Europeans and the Indians:—“The natives of
the island, when at the dawn of day they had beheld the ships
hovering on the coast, had supposed them some monsters, which
had issued from the deep during the night. When they beheld the
boats approach the shore, and a number of strange beings, clad in
glittering steel, or raiment of various colors, landing upon the beach,
they fled in affright to the woods.
“Finding, however, that there was no attempt to pursue or molest
them, they gradually recovered from their terror, and approached the
Spaniards with great awe, frequently prostrating themselves, and
making signs of adoration. During the ceremony of taking
possession, they remained gazing, in timid admiration, at the
complexion, the beards, the shining armor, and splendid dress of the
Spaniards.
Columbus landing.
“The admiral particularly attracted their attention, from his
commanding height, his air of authority, his scarlet dress, and the
deference paid him by his companions; all which pointed him out to
be the commander.
“When they had still further recovered from their fears, they
approached the Spaniards, touched their beards, and examined their
hands and faces, admiring their whiteness. Columbus was pleased
with their simplicity, their gentleness, and the confidence they
reposed in beings who must have appeared so strange and
formidable, and he submitted to their scrutiny with perfect
acquiescence.
“The wondering savages were won by this benignity. They now
supposed that the ships had sailed out of the crystal firmament
which bounded their horizon or that they had descended from above,
on their ample wings, and that these marvellous beings were
inhabitants of the skies.
“The natives of the island were no less objects of curiosity to the
Spaniards, differing, as they did, from any race of men they had
seen. They were entirely naked, and painted with a variety of colors
and devices, so as to give them a wild and fantastic appearance.
Their natural complexion was of a tawny or copper hue, and they
had no beards. Their hair was straight and coarse; their features,
though disfigured by paint, were agreeable; they had lofty foreheads,
and remarkably fine eyes.
“They were of moderate stature, and well shaped. They appeared
to be a simple and artless people, and of gentle and friendly
dispositions. Their only arms were lances, hardened at the end by
fire, or pointed with a flint or the bone of a fish. There was no iron
among them, nor did they know its properties, for when a drawn
sword was presented to them, they unguardedly took it by the edge.
“Columbus distributed among them colored caps, glass beads,
hawk’s bells, and other trifles, which they received as inestimable
gifts, and decorating themselves with them, were wonderfully
delighted with their finery. In return, they brought cakes of a kind of
bread called cassava, made from the yuca root, which constituted a
principal part of their food.”
Columbus distributing presents.

Thus kindly began the intercourse between the Old World and the
New; but the demon of avarice soon disturbed their peace. The
Spaniards perceived small ornaments of gold in the noses of some
of the natives. On being asked where this precious metal was
procured, they answered by signs, pointing to the south, and
Columbus understood them to say that a king resided in that quarter,
of such wealth that he was served in great vessels of gold.
Columbus took seven of the Indians with him, to serve as
interpreters and guides, and set sail to find the country of gold. He
cruised among the beautiful islands, and stopped at three of them.
These were green, fertile, and abounding with spices and
odoriferous trees. The inhabitants, everywhere, appeared the same
—simple, harmless, and happy, and totally unacquainted with
civilized man.
Columbus was disappointed in his hopes of finding any gold or
spices in these islands; but the natives continued to point to the
south, and then spoke of an island in that direction called Cuba,
which the Spaniards understood them to say abounded in gold,
pearls, and spices. People often believe what they earnestly wish;
and Columbus sailed in search of Cuba, fully confident that he
should find the land of riches. He arrived in sight of it on the 28th of
October, 1492.
Here he found a most lovely country, and the houses of the
Indians, neatly built of the branches of palm trees, in the shape of
pavilions, were scattered under the trees, like tents in a camp. But
hearing of a province in the centre of the island, where, as he
understood the Indians to say, a great prince ruled, Columbus
determined to send a present to him, and one of his letters of
recommendation from the king and queen of Spain.
For this purpose he chose two Spaniards, one of whom was a
converted Jew, and knew Hebrew, Chaldaic, and Arabic. Columbus
thought the prince must understand one or the other of these
languages. Two Indians were sent with them as guides. They were
furnished with strings of beads, and various trinkets, for their
travelling expenses, and they were enjoined to ascertain the
situation of the provinces and rivers of Asia, for Columbus thought
the West Indies were a part of the Eastern Continent.
The Jew found his Hebrew, Chaldaic, and Arabic of no avail, and
the Indian interpreter had to be the orator. He made a regular speech
after the Indian manner, extolling the power, wealth, and generosity
of the White men. When he had finished, the Indians crowded round
the Spaniards, touched and examined their skin and raiment, and
kissed their hands and feet in token of adoration. But they had no
gold to give them.
It was here that tobacco was first discovered. When the envoys
were on their return, they saw several of the natives going about with
firebrands in their hands, and certain dried herbs which they rolled
up in a leaf, and lighting one end, put the other in their mouths, and
continued inhaling and puffing out the smoke. A roll of this kind they
called a tobacco. The Spaniards were struck with astonishment at
this smoking.
Indians smoking.
When Columbus became convinced that there was no gold of
consequence to be found in Cuba, he sailed in quest of some richer
lands, and soon discovered the island of Hispaniola, or Hayti. It was
a beautiful island. The high mountains swept down into luxuriant
plains and green savannas, while the appearance of cultivated fields,
with the numerous fires at night, and the volumes of smoke which
rose in various parts by day, all showed it to be populous. Columbus
immediately stood in towards the land, to the great consternation of
his Indian guides, who assured him by signs that the inhabitants had
but one eye, and were fierce and cruel cannibals.
(To be continued.)
Shocking.—An Irish carman and his wife attended the wake, on

Friday night, over the body of John Hand, whom Cliff killed. To do so,
they left twin infants, fourteen months old, in the cradle at home; but,
becoming intoxicated, they did not return until morning, when they
found their infants dead! The decision of the coroners’ jury was, we
understand, that they came to their death by cold and starvation.—
Detroit Adv., 1840.
View of St. Paul’s Bay, Malta.
The Travels, Adventures, and Experiences of
Thomas Trotter.
CHAPTER IV.
Landing at Malta.—Description of the city and inhabitants.—
Excursion into the interior.—Visit to the catacombs.—
Wonderful subterranean abodes.—St. Paul’s Bay.
When we were through with the quarantine, we hauled round into

the great harbor of Malta. The city, which is called Valetta, made a
most stately appearance as we passed the castle of St. Elmo. It lies
close to the sea, and the whole mass of buildings bursts upon you at
once, with its long rows of castellated walls, bristling with cannon,
tier upon tier, towering battlements, turrets and bastions and
pinnacles in the most picturesque profusion—a grand and
magnificent spectacle. The harbor was full of ships—men-of-war,
merchantmen, and all sorts of small Mediterranean craft, rigged in
the strangest style imaginable.
Whole fleets of row-boats came crowding round us, filled with
people. Some of them had bands of music, playing “Yankee Doodle,”
“Washington’s March,” and “Hail Columbia,” for which they expected
we should give them a quarter of a dollar or so. Others brought fruit,
fresh provisions, sea-shells and curiosities, for sale. Most of them
spoke a little English, and, in their eagerness to sell their
commodities, would make the most ludicrous speeches imaginable.
One comical fellow had a pig for sale, which he praised very highly:
“Buy pig, captain?—nice pig, sweet pig, ’merican pig: won’t heave
nothing overboard, eat brick-dust, eat anything.” It was difficult to get
rid of the importunities of these people. They would offer a thing for a
dollar, and then gradually come down to nine-pence.
When we landed on the quay, we found a still greater crowd
besetting us, offering to carry our trunks, amidst immense confusion
and jabbering. Donkeys and mules were trotting about, but we saw
no horses. We passed through a great gate in a wall, and went up
into the city, by climbing flights of stone steps. The donkeys go up
and down with heavy loads on their backs, and never stumble. All
the streets were narrow, with high stone houses on each side, and
full of people. The main street occupies the summit of the rock on
which the city is built, and all the cross streets run up and down the
hill, and are paved stair-fashion. The city is one of the handsomest in
the world, and looks like an assemblage of palaces. The streets are
straight, and all the houses are built of a light yellow stone. Nothing
can be more picturesque than their architecture. The fronts are
studded with bold masses of carved stonework, balconies, cornices,
pilasters, projections, and sculptured ornaments of various
descriptions. The prospect through one of the streets is a perfect
picture. I could not help contrasting it with our American cities, with
their quadrangular monotony of architecture!
After we had secured our baggage at the hotel, I walked out to
take a view of the city. The population seemed to be all in the
streets, and to live out of doors. The crowd was immense in every
public place, and everything visible was full of character and variety.
I do not believe there is a spot in the world that exhibits a more
striking and motley spectacle than the streets of Malta. This island is
the central point of the whole Mediterranean commerce, which
brings it a constant succession of visitors from all the countries
around. The crowd looks like a fancy ball, where the people dress so
as to differ from each other. Here is the fantastical Greek in his
picturesque drapery of red and white; the turbaned Turk with his
bushy and flowing beard; the swarthy Arab in his coarse haick or
cloak; the grave Austrian, the scowling Moor, the squab Dutchman,
the capering Sicilian, the hawk-eyed and tawny Calabrian, the native
Maltese; the Spaniard, the Frenchman, the John Bull, and the
Yankee, all in strange mixture, and with their various manners and
languages. The whole group is perfectly dramatic. Little boys, about
as high as my knee, were running about, dressed in black small-
clothes and those great cocked-hats which we call “three-cornered
scrapers.” The women of the island looked like nuns in black silk
hoods; they cover most of the face, and peep out with one eye. This
habit makes almost all the women squint-eyed.
After I had gone over the greater part of the city and visited its
elegant churches, of which it contains a large number, I set forth for
a walk into the country. I went out at a massive gateway and across
a draw-bridge, which offer the only passage-way into the interior of
the island. I was struck with astonishment at the strength and extent
of the fortifications. It seemed impossible that any force, either of
human arms or cannon-balls, could ever break through the walls.
The French took the place in 1800, and when Bonaparte entered at
this gate, he said it was lucky there was somebody inside to open it,
or they could never have got in. Immense walls and bastions, one
above another, towered over my head. I looked down into one of the
ditches; it appeared to be a hundred feet deep, and there were
flower-gardens and orchards at the bottom. After travelling a few
minutes, I saw before me a long row of arches, fifteen or twenty feet
high, which I found to be an aqueduct: the road passed under it.
Here I had the first glimpse of the country, and I was struck with the
odd appearance of everything. There were no fields nor pastures,
such as we have in our country, but the whole land lay in terraces,
faced with thick stone walls, making little square inclosures, where
crops of wheat and other vegetables were growing. The whole face
of the island presented a succession of stripes of light yellow rock
and fresh green vegetation. Here and there were low hills dotted with
dark green locust trees, and a great many country houses and villas
were scattered round. All along under these walls grew wild fig-trees
and immense clumps of prickly pear, and thousands of lizards were
darting up and down with the liveliest movements. Peasants were
passing along the road driving donkeys loaded with bundles of
grass, and now and then I met a chaise drawn by a mule, thumping
over the stony road. I was surprised that any person could be found
willing to risk his bones by such a jolting.
One would suppose, by the looks of the country here, that the
inhabitants had covered it with stone walls to keep the grass from
blowing away. Indeed, the soil is so thin, and the surface so irregular,
that but for these walls, half the island would be washed bare by the
rains. It is a solid rock, with only a foot or two of soil. Having gone
several miles, I reached Citta Vecchia, or the old city, the ancient
capital of Malta. It stands in the centre of the island, and looks very
antique, being a confused assemblage of fantastical structures, gray
with age. It is probably three thousand years old or more. I went into
a little shop kept by an old woman, and amused myself with staring
at the odd appearance of everything. A man sat at work cutting a
pair of sandals out of a raw hide; a little boy, with a desperately dirty
face, was munching a handful of green stuff in a corner; and a
queer-looking blue cat, with half a tail, rolled her green eyes up at
me: she had doubtless never seen a Yankee before. The old woman
sold bread, greens, oranges, wine, &c. I drank a tumbler of wine, for
which I paid a half-penny; it was a dark red wine like claret, and
about the strength of common cider. Some wine is made in the
island, but most of what is used comes from Sicily.
I went to the top of the great church, which has a very lofty dome,
where I had a prospect of the whole island. The view is picturesque
and striking in the highest degree. The island looks like an immense
chess-board, the surface being chequered out into squares of green
verdure and stone wall. Villages without number were scattered
about in every direction, each with the tall dome of a church rising
above its cluster of houses. Many of these churches I visited in my
walk, and was astonished to find every one of them richly adorned in
the interior with gold, silver, and precious stones. The private houses
in these villages are very far from exhibiting the same wealth.
I had a guide with me, who showed me over the cathedral of Citta
Vecchia, and then asked if I wanted to see the catacombs. I had
never before heard of them, but replied in the affirmative; whereupon
he led the way through a narrow street, till he came to a door, at
which he thumped lustily. It was opened by a little tawny-faced fellow
in a monk’s dress. He bustled about and got a bunch of keys, and
some torches and candles. We each took a torch and candle, and
followed him through a series of long narrow lanes, till we came to a
great gate in a wall. Here we struck fire, lighted our torches and
candles, and entered the passage. It looked dark and dismal, and we
continued going down long flights of steps till we came to a sort of
landing-place at a great distance below the surface. I know not how
to describe the scene that I witnessed here. For miles around, there
was a labyrinthian extent of dark passages cut in the rock, winding
and zigzagging in all directions; sometimes expanding into the
breadth and loftiness of spacious halls, and sometimes contracting
into a strait so narrow as hardly to admit a single person.
Along the sides of these galleries were innumerable niches and
recesses cut in the rock as places of deposit for corpses; they were
probably all full, thousands of years ago. Here and there we found a
solitary bone, which I gazed at with feelings of awe as the relic of an
ancient generation. The place appeared to me like a great
subterranean city whose inhabitants had all deserted it. The age of it
is unknown; not even tradition can tell it. It was used as a hiding-
place by the early Christians during times of persecution, and must
have been found admirably suited to that purpose: thousands and
thousands of people might conceal themselves beyond all search in
its immense extent of winding and perplexing avenues, which run
into one another, and would lead any one astray who was not
perfectly familiar with all their turnings and windings.
In one of the large halls we found two ancient hand-mills for corn
and oil, which had been used by the inhabitants of this dark abode.
Every passage and room is full of secret nooks and openings, into
which the inmates might creep for safety in case of surprise. Great
numbers of names and inscriptions are cut in all parts of the rock;
and the sides and ceiling of the narrow galleries are blackened with
the smoke of torches. Strange and overpowering were the
sensations that came upon me as I followed my guide through these
drear avenues and halls of death. In spite of my confidence in him, it
was impossible not to feel an apprehension of being lost among the
innumerable turnings and windings of this dark labyrinth. Now and
then we would stop and contemplate the striking effect of our
flickering torches, which threw red gleams of light along the walls,
and seemed to show us indistinct forms flitting hither and thither
amid the darkness beyond.
We stood still, held our breath, and marked the drear silence that
reigned around, where the sound of a footstep or a whisper struck
the ear like an unhallowed intrusion breaking the still repose of the
ancient dead. Then we shouted and listened to the hollow echoes
that rumbled through the rocky mansion, and died away in the
distance, among miles of long galleries and reverberating caverns.
No scene could be more impressive—I almost expected the dead
inmates of this gloomy abode to start up before my face, and greet
me with the accents of three thousand years ago. We traversed one
long passage after another, but the labyrinth appeared to be endless.
The excavations are said to be fifteen miles in extent; they may be
twice as long for aught I know: the only wonder is that any man ever
undertook to measure them. After all I have said, the reader will have
no adequate conception of these wonderful abodes: he must go to
the spot to know what they really are.
I never knew the light of day so cheerful, delicious, and
exhilarating as when I got out of this dark place, into the open air; it
seemed like passing from death to life. The little monk was very
thankful for a ninepence which I gave him for his trouble in showing
me through the catacombs.
Going along one of the streets of the town, I saw a statue of St.
Paul, shaking the viper from his hand. This is believed to be the spot
where the house stood in which he lodged while in the island. There
is a bay on the southwestern shore, where, according to tradition, he
was shipwrecked. This I determined to visit, and hired a stout boy,
whom I found in the street, to show me the way. We travelled over a
road on the bare rock, very rough, and which grew rougher every
mile. The country was pretty much like what I have mentioned,
parcelled out into little square inclosures, with low cabins in the sides
of the walls, looking like dog-kennels, but designed as lodging-
places for the men who guard the fields by night. By-and-by the road
began to descend, and I soon found we were close to the sea. I was
obliged to clamber down the ragged rocks, but my companion
jumped from cliff to cliff like a goat. We soon reached the margin of
the bay, and he conducted me to a bold projection in the rocky
shore, which tradition has marked out as the precise spot where the
ship which was bearing St. Paul to Rome, struck the land, as related
in the twenty-seventh chapter of the Acts of the Apostles.
I walked out to the extremity of the point, against which the sea
was dashing, and sat down upon the rock to enjoy the feelings
excited by the history of this interesting place. I gazed for some time
upon the wild scene around me, and called up in imagination the
shadows of the beings who, 1800 years ago, had figured in these
events. Here stood the shipwrecked apostle and beheld the same
wild and rugged prospect that strikes the eye at the present moment,
for hardly a single point in the landscape appears to have undergone
any change since his time. There is a chapel on the shore a few
yards from the water, and two or three castles on the eminences
around; these are all the buildings in sight. Three or four ragged
boys were picking up shells on the beach, but no other living
creature was to be seen. I saw the sun sink into the ocean, and was
obliged to hasten my return, lest the city gates should be closed.
(To be continued.)
Wit.—Some one observed to a wag on one occasion, that his

coat seemed to have been made too short; to which he replied, that
“it would be long enough before he got another.”
In delay, there lies no plenty.

The Kingfisher and the Nightingale;
a fable.
Once upon a time, a meeting took place between a kingfisher and

a mocking-bird. The latter, being dressed in very plain feathers, at
first felt a little humbled by the brilliant plumage of his neighbor. The
kingfisher, perceiving the admiration of the mocking-bird, jerked his
tail and tossed his head, so as to show off all the changing hues of
his feathers to great advantage.
While this was going on between the two birds, a sportsman
chanced to be passing by, and seeing them, paused to watch their
proceedings. Readily understanding the scene, and disgusted with
the conceit and vanity of the kingfisher, he drew up his gun, and shot
him down. As he went to pick up the fallen bird, he made the
following reflections:
“This silly kingfisher is like a person who is vain of his dress or his
outward beauty. His skin, when stuffed with tow, is just as valuable
as when the bird’s living flesh and bones are in it; his outside is all
there is of him. But the modest mocking-bird is like a person who
contributes to our pleasure or our instruction, and relies upon the
good he does to others for his standing among mankind. How
contemptible is pride; how amiable and attractive is modesty allied to
merit!”
A sagacious Dog.—A grocer in Edinburgh had a dog, which for

some time amused and astonished the people in the neighborhood.
A man who went through the streets ringing a bell and selling penny
pies, happened one day to treat this dog with a pie. The next time he
heard the pieman’s bell, the dog ran to him with impetuosity, seized
him by the coat, and would not allow him to pass. The pieman, who
understood what the animal wanted, showed him a penny, and
pointed to his master, who stood at the street door and saw what
was going on. The dog immediately supplicated his master by many
humble gestures and looks. The master put a penny into the dog’s
mouth, which he instantly delivered to the pieman, and received his
pie; and this traffic between the pieman and the grocer’s dog
continued to be daily practised for many months.
Absence of Mind.
This is that habit which some people have, of thinking of one

thing, while they are doing another. The famous Sir Isaac Newton
was a philosopher, and he thought a great deal about the heavenly
bodies, and such mighty matters. Of course, he could hardly be
expected to think much about common things. However, he did once
have a fancy for a lady, and one evening he went to see her. As he
was sitting with her by the fireside smoking his pipe, he became
absorbed in his mathematics, and in his absence of mind he took
hold of the lady’s finger and stuck it into the fiery bowl of his pipe,
thus making it a tobacco-stopper!
I once knew an old lady who would go about the room, looking
upon the shelf, peeping into the table drawer, tumbling over a
cupboard that served as a kind of Noah’s ark, where every strange
thing was deposited—all the time teasing and fretting because she
could not find her spectacles, until at last she discovered that the
said spectacles were snugly sitting astride of her nose!
But this is a trifling instance of absence of mind, compared with
some others. An old maid of Edinburgh, in Scotland, had taken an
unaccountable fancy to a pig, which she kept as a kind of pet about
the house, and often took it into her lap. The poor thing seemed to
be forever pinched with a pain in its bowels, and therefore kept up an
almost perpetual squealing. Still, the kind woman loved it all the
better, and cherished it the more for its very infirmities. The lady was
withal a literary lady, and fond of reading and writing books, and her
head ran upon these operations so much, that she often forgot
where she was, and what she was doing.
One day, she appeared at the door of a neighbor in a good deal of
trouble, with the pig under her arm, squealing with all its might, as
usual; upon which the following dialogue ensued:
Woman. Good morning, neighbor! Good morning! I called to see
you about—about—something or other—but in fact I forget what it
was I was after.
Neighbor. Oh! you wanted something or other, and you thought
you’d come and ask me what ’twas you wanted?
Woman. Why yes—no. Be still, you naughty pig! be still! Yes, I am
looking for something. Stop your everlasting squealing! Oh! I
remember! I’ve lost my pig. Have you seen anything of him?
Neighbor. Why, what’s that you have under your arm?
Woman. Gracious! I’ve got the pig under my arm all this time!
Poor, dear thing—that I should have forgotten you, while I was all the
time thinking of you! and that I should have lost you while I was
clasping you to my breast! Well done! I must be a genius, as aunt
Dorcas says!
Some years ago there lived at the city of Washington a famous
Englishman by the name of Thomas Law. He was very absent-
minded, and often forgot his christian name. One day, he was writing
a letter, and when he came to the end, and wanted to sign his name,
he was in great trouble because he could not remember the first part
of it. At last, Claxton, the door-keeper, chanced to be passing, and
Law remembered that his christian name was the same as Claxton’s.
Accordingly he said, “Claxton, what is your christian name?”
“Thomas,” was the answer. “Oh yes, Thomas,” said Law, and
immediately wrote his name, “Thomas Law!”
These instances are somewhat amusing, but I can tell you of an
instance in which absence of mind proved more serious. A famous
courtier once wished to ingratiate himself into the favor of two
persons of great rank and power, but who were deadly enemies to
each other. These were Lord B. and Lord Q. In order to please these
two persons, the courtier wrote a letter to each of them. That of Lord
B. was as follows:
My dear Lord B.
I met with Lord Q. last evening at Lady Lackaday’s. It was
the first time I had seen him. I felt instinctively an aversion
similar to that which is inspired by the presence of a serpent. I
can easily enter into your feelings respecting him. Indeed, I do
not see how any one can differ from your lordship in this
matter. It is impossible not to feel a sympathy with the man
who stands in open and manly opposition to one upon whose
forehead “knave” is written by the hand of his Creator.
I am, dear Lord, yours,
B. L.
The next letter was as follows:
My dear Lord Q.
Lord B. is an ass, and I ask no better proof of it than that
he seems to hate you, whom all the world beside agree to
love and admire. He is stark mad with envy. You have only to
let him alone, and he will make himself ridiculous before the
whole town. This is all you have to do to destroy your rival.
Let him alone! Yours faithfully,
B. L.
Such are the two letters; but unluckily for the success of the
courtier’s crafty schemes, he was addicted to fits of absence of mind,
and when he came to superscribe the aforesaid letters, he
addressed the one intended for Lord B. to Lord Q., and that for Lord
Q. to Lord B.; so that when they were read, each of these persons
discovered the trick and hypocrisy of the courtier.

Precalculus Mathematics For Calculus International Metric Edition 7th Edition Stewart Solutions Manual Instant Download All Chapter

Uploaded by

Copyright:

Available Formats

Precalculus Mathematics For Calculus International Metric Edition 7th Edition Stewart Solutions Manual Instant Download All Chapter

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Precalculus Mathematics For Calculus International Metric Edition 7th Edition Stewart Solutions Manual Instant Download All Chapter

Uploaded by

Copyright:

Available Formats

Precalculus Mathematics for Calculus

International Metric Edition 7th Edition

Precalculus Mathematics for Calculus 7th Edition

Precalculus Mathematics for Calculus 7th Edition

Precalculus Mathematics for Calculus 6th Edition

Calculus 7th Edition Stewart Solutions Manual

Calculus An Applied Approach Brief International Metric

Calculus Early Transcendentals 7th Edition Stewart

Understanding Basic Statistics International Metric

Calculus 7th Edition Stewart Test Bank

6.1 ANGLE MEASURE

 rad   7 rad s 1222 rad

33.    7  15  9  17

triangle is a right triangle. Therefore,   

6.2 TRIGONOMETRY OF RIGHT TRIANGLES

(a) sin   cos   47 (b) tan   cot   T4 (c) sec   csc   T7

11. (a) sin 22i s 037461 (b) cot 23i s 235585

12. (a) cos 37i s 079864 (b) csc 48i s 134563

13. (a) sec 13i s 102630 (b) tan 51i s 123490

14. (a) csc 10i s 575877 (b) sin 46i s 071934

6.3 TRIGONOMETRIC FUNCTIONS OF ANGLES

(b) The reference angle for 98 is 98    

(c) The reference angle for 103 is 103  3  

(c) The reference angle for 14 is 14 because 14  

25. sin 32   sin 

37. Since sin   0 and cos   0,  is in quadrant III.

57. The area is 12 7 9 sin 72i s 300.

From the graph, the largest area is achieved when

72. (a) We label the lengths L 1 and L 2 as shown in the (b)

0 n a n 180 and sin a  b, then either a  b or 180  a  b. The   Y

graph shows that the first positive solution occurs when

6.4 INVERSE TRIGONOMETRIC FUNCTIONS AND TRIANGLES

6. (a) sin1 0  0 (b) cos1 1   (c) tan1 0  0

21. sin   47 , so   sin1 47 s 348i . 22. cos   89 , so   cos1 89 s 273i .

41. (a) Solving tan   h2 for h, we have h  2 tan .

43. (a) Solving sin   h204 for  we have   sin1 h204.

6.5 THE LAW OF SINES

36. % The situation is illustrated in the diagram.

6.6 THE LAW OF COSINES

31. The semiperimeter is s  789 2  12, so by Heron’s Formula the area is

33. The semiperimeter is s  346 2  13

9 sin 7053i s 077139 " C s sin1 077139 s 5048i . The area of

ABC is 12 AB AC sin A  12 9 10 sin 7053i  s 42426.

/ B  / D  360i  100i  260i . Thus, / B  / D 

x 2  302  502  2  30  50  cos 100i

x  292 5  2402  10202 s 1151 m.

3. (a) 56 rad  56  180

(c)  43 rad   43  180

(d) 4 rad  4  180 720

5. r  10 m,   25 rad. Then s  r  10  25  4 s 126 m.

11. r  5 m,   2 rad. Then A  12 r 2   12  52  2  25 m2 .

30. As the crankshaft moves in its circular pattern, point Q is Z Z

determined by the angle , namely it has coordinates

2. 43 rad  43  180 i 180 234

5 , tan    5 . Then sec   1 1 sin  1 5 13 13

FOCUS ON MODELING Surveying

In the bank-second bridge-cliff triangle, the third angle is 

Finally, in the school-fire hall-cliff triangle, the third angle is c

Columbus distributing presents.

rad 7 rad s 1222 rad

33. 7 15 9 17

triangle is a right triangle. Therefore,

(a) sin cos 47 (b) tan cot T4 (c) sec csc T7

11. (a) sin 22i s 037461 (b) cot 23i s 235585

12. (a) cos 37i s 079864 (b) csc 48i s 134563

13. (a) sec 13i s 102630 (b) tan 51i s 123490

14. (a) csc 10i s 575877 (b) sin 46i s 071934

(b) The reference angle for 98 is 98

(c) The reference angle for 103 is 103 3

(c) The reference angle for 14 is 14 because 14

25. sin 32 sin

37. Since sin 0 and cos 0, is in quadrant III.

57. The area is 12 7 9 sin 72i s 300.

0 n a n 180 and sin a b, then either a b or 180 a b. The Y

6. (a) sin1 0 0 (b) cos1 1 (c) tan1 0 0

21. sin 47 , so sin1 47 s 348i . 22. cos 89 , so cos1 89 s 273i .

41. (a) Solving tan h2 for h, we have h 2 tan .

43. (a) Solving sin h204 for we have sin1 h204.

31. The semiperimeter is s 789 2 12, so by Heron’s Formula the area is

33. The semiperimeter is s 346 2 13

9 sin 7053i s 077139 " C s sin1 077139 s 5048i . The area of

ABC is 12 AB AC sin A 12 9 10 sin 7053i s 42426.

/ B / D 360i 100i 260i . Thus, / B / D

x 2 302 502 2 30 50 cos 100i

x 292 5 2402 10202 s 1151 m.

3. (a) 56 rad 56 180

(c) 43 rad 43 180

(d) 4 rad 4 180 720

5. r 10 m, 25 rad. Then s r 10 25 4 s 126 m.

11. r 5 m, 2 rad. Then A 12 r 2 12 52 2 25 m2 .

determined by the angle , namely it has coordinates

2. 43 rad 43 180 i 180 234

5 , tan 5 . Then sec 1 1 sin 1 5 13 13

In the bank-second bridge-cliff triangle, the third angle is

Finally, in the school-fire hall-cliff triangle, the third angle is c