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Chapter 7
Additional Topics in Trigonometry
Section 7.1
a b
Check Point Exercises 
sin A sin B
a 12
1. Begin by finding B, the third angle of the triangle. 
A  B  C  180 sin 40 sin117.5
64  B  82  180 12 sin 40
a  8.7
sin117.5
146  B  180
Use the Law of Sines again, this time to
B  34 find c.
In this problem, we are given c and C: c b
c = 14 and C = 82°. Thus, use the ratio 
sin C sin B
c 14
, or , to find the other two sides. Use c 12
sin C sin 82 
the Law of Sines to find a. sin 22.5 sin117.5
12 sin 22.5
a c c  5.2
 sin117.5
sin A sin C
The solution is B = 117.5º, a ≈ 8.7, and c ≈ 5.2.
a 14

sin 64 sin 82 a 33
3. The known ratio is , or . Because side b
14 sin 64
a sin A sin 57
sin 82 is given, Use the Law of Sines to find angle B.
a  12.7 centimeters a b
Use the Law of Sines again, this time to find b. 
sin A sin B
b c 33 26
 
sin B sin C sin 57 sin B
b 14 33sin B  26sin 57

sin 34 sin 82 26sin 57
14 sin 34 sin B   0.6608
b 33
sin 82 sin B  0.6608
b  7.4 centimeters
B  41
The solution is B = 34º, a ≈ 12.7 centimeters, and b ≈
7.4 centimeters. 180  41  139 also has this sine value, but, the
sum of 57 and 139 exceeds 180, so B cannot have
2. Begin by finding B. this value.
A  B  C  180 C  180  B  A  180  41  57  82 .
40  B  22.5  180 Use the law of sines to find C.
62.5  B  180 a c

B  117.5 sin A sin C
In this problem, we are given that b = 12 and we find 33 c

that B = 117.5°. Thus, use the ratio sin 57 sin 82
b 12 33sin 82
, or , to find the other two sides. Use c
sin B sin117.5 sin 57
the Law of Sines to find a. c  39
Thus, B  41, C  82, c  39.
Copyright © 2014 Pearson Education, Inc. 823

Chapter 7 Additional Topics in Trigonometry
a 10 6. The area of the triangle is half the product of the

sin A sin 50 lengths of the two sides times the sine of the included
is given, Use the Law of Sines to find angle B. angle.
1
a

b Area  (8)(12)(sin135)  34
sin A sin B 2
The area of the triangle is approximately 34 square
10 20
 meters.
sin 50 sin B
10sin B  20sin 50 7.
20sin 50
sin B   1.53
10
Because the sine can never exceed 1, there is no
angle B for which sin B  1.53 . There is no triangle
with the given measurements.
a 12
sin A sin 35
is given, Use the Law of Sines to find angle B. Using a north-south line, the interior angles are found
a b as follows:
 A  90  35  55
sin A sin B
12 16 B  90  49  41
 Find angle C using a 180° angle sum in the triangle.
sin 35 sin B C  180  A  B  180  55  41  84
12 sin B  16sin 35
c 13
16sin 35 The ratio , or
sin 84
is now known. Use this
sin B   0.7648 sin C
12 ratio and the Law of Sines to find a.
There are two angles possible: a c
B1  50, B2  180  50  130 
sin A sin C
There are two triangles: a 13
C1  180  A  B1  180  35  50  95 
sin 55 sin 84
C2  180  A  B2  180  35  130  15 13sin 55
a  11
Use the Law of Sines to find c1 and c2 . sin 84
c1 a The fire is approximately 11 miles from station B.

sin C1 sin A
c1 12
 Concept and Vocabulary Check 7.1
sin 95 sin 35
12sin 95
c1   20.8 1. oblique; sides; angles
sin 35
c2 a a b c
 2.  
sin C2 sin A sin A sin B sin C
c2 12
 3. side; angles
sin15 sin 35
12sin15 4. false
c2   5.4
sin 35
1
5. ab sin C
In one triangle, the solution is B1  50 , 2
C1  95, and c1  20.8 . In the other triangle,
B2  130, C2  15, and c2  5.4 .
824 Copyright © 2014 Pearson Education, Inc.

Section 7.1 The Law of Sines
Exercise Set 7.1 Use the Law of Sines again, this time to find b.
b c
1. Begin by finding B. 
sin B sin C
A  B  C  180
b 12
42  B  96  180 
sin 48 sin 90
138  B  180 12sin 48
b
B  42 sin 90
Use the ratio
c
, or
12
, to find the other two b  8.9
sin C sin 96 The solution is C  90, a  8.0, and b  8.9 .
sides. Use the Law of Sines to
find a. 3. Begin by finding A.
a c A  B  C  180

sin A sin C A  54  82  180
a 12 A  136  180

sin 42 sin 96 A  44
12sin 42
a Use the ratio
a
, or
16
, to find the other two
sin 96 sin A sin 44
a  8.1 sides. Use the Law of Sines to
Use the Law of Sines again, this time to find b.
find b. b a
b c 
 sin B sin A
sin B sin C b 16
b 12 

 sin 54 sin 44
sin 42 sin 96
16sin 54
12sin 42 b
b sin 44
sin 96
b  8.1 b  18.6
The solution is B  42, a  8.1, and b  8.1 .
Use the Law of Sines again, this time to
find c.
2. Begin by finding C.
c a
A  B  C  180 
sin C sin A
42  48  C  180
c 16
90  C  180 
sin 82 sin 44
C  90 16sin 82
c 12 c
Use the ratio , or , to find the other two sin 44
sin C sin 90º c  22.8
sides. Use the Law of Sines to find a. The solution is A  44, b  18.6, and
a c
 c  22.8 .
sin A sin C
a 12

sin 42 sin 90
12sin 42
a
sin 90
a  8.0

4. Begin by finding B. Use the Law of Sines again, this time to find c.
A  B  C  180 c a

33  B  128  180 sin C sin A
B  161  180 c

100
B  19 sin 95 sin 48
100sin 95
Use the ratio
a
, or
16
, to find the other two c
sin A sin 33 sin 48
sides. Use the Law of Sines to find b. c  134.1
b a The solution is C  95, b  81.0, and c  134.1 .

sin B sin A
b 16
 A  B  C  180
sin19 sin 33
6  12  C  180
16sin19
b 18  C  180
sin 33
b  9.6 C  162
Use the Law of Sines again, this time to find c. c 100
Use the ratio , or , to find the other
c a sin C sin162

sin C sin A two sides. Use the Law of Sines to find a.
c 16 a c
 
sin128 sin 33 sin A sin C
16sin128 a

100
c
sin 33 sin 6 sin162
c  23.1 100sin 6
a
The solution is B  19, b  9.6, and c  23.1 . sin162
a  33.8
5. Begin by finding C. Use the Law of Sines again, this time to find b.
A  B  C  180 b c

48  37  C  180 sin B sin C
85  C  180 b

100
C  95 sin12 sin162
100sin12
Use the ratio
a
, or
100
, to find the other two b
sin A sin 48 sin162
sides. Use the Law of Sines to find b. b  67.3
b a The solution is C  162, a  33.8, and b  67.3 .

sin B sin A
b 100

sin 37 sin 48
100sin 37
b
sin 48
b  81.0

A  B  C  180 c a

38  B  102  180 sin C sin A
B  140  180 c

20
B  40 sin 40 sin 38
20sin 40
Use the ratio
a
, or
20
b a The solution is C  40, b  31.8, and c  20.9 .

sin B sin A
b 20
 A  B  C  180
sin 40 sin 38
44  25  C  180
20sin 40
b 69  C  180
sin 38
b  20.9 C  111
Use the Law of Sines again, this time to find c. a 12
Use the ratio , or , to find the other two
c a sin A sin 44
 sides. Use the Law of Sines to find b.
sin C sin A
c 20 b a
 
sin102 sin 38 sin B sin A
20sin102 b 12
c 
sin 38 sin 25 sin 44
c  31.8 12sin 25
b
The solution is B  40, b  20.9, and c  31.8 . sin 44
b  7.3
A  B  C  180 Use the Law of Sines again, this time to find c.
38  102  C  180 c a

140  C  180 sin C sin A
c 12
C  40 
sin111 sin 44
a 20
Use the ratio , or , to find the other two 12sin111
sin A sin 38 c
sides. Use the Law of Sines to find b. sin 44
b a c  16.1
 The solution is C  111, b  7.3, and c  16.1 .
sin B sin A
b 20

sin102 sin 38
20sin102
b
sin 38
b  31.8

A  B  C  180 c b

56  B  24  180 sin C sin B
B  80  180 c

40
B  100 sin15 sin 85
40sin15
Use the ratio
a
, or
22
b a The solution is A  80, a  39.5, and c  10.4 .

sin B sin A
b 22
 A  B  C  180
sin100 sin 56
85  35  C  180
22sin100
b 120  C  180
sin 56
b  26.1 C  60
Use the Law of Sines again, this time to find c. c 30
c a sin C sin 60

sin C sin A sides. Use the Law of Sines to find a.
c 22 a c
 
sin 24 sin 56 sin A sin C
22sin 24 a

30
c
sin 56 sin 85 sin 60
c  10.8 30sin 85
a
The solution is B  100, b  26.1, and c  10.8 . sin 60
a  34.5
11. Begin by finding A. Use the Law of Sines again, this time to find b.
A  B  C  180 b c

A  85  15  180 sin B sin C
A  100  180 b

30
A  80 sin 35 sin 60
30sin 35
Use the ratio
b
, or
40
, to find the other two b
sin B sin 85 sin 60
sides. Use the Law of Sines to find a. b  19.9
a b The solution is C  60, a  34.5, and b  19.9 .

sin A sin B
a 40

sin 80 sin 85
40sin 80
a
sin 85
a  39.5

A  B  C  180 c b

115  B  35  180 sin C sin B
B  150  180 c

200
B  30 sin125 sin 5
200sin125
Use the ratio
c
, or
200
sin C sin 35 sin 5
sides. Use the Law of Sines to find a. c  1879.7
a c The solution is A  50, a  1757.9, and c  1879.7 .

sin A sin C
a 200
 A  B  C  180
sin115 sin 35
65  65  C  180
200sin115
a 130  C  180
sin 35
a  316.0 C  50
Use the Law of Sines again, this time to find b. c 6
b c sin C sin 50

sin B sin C sides. Use the Law of Sines to find a.
b 200 a c
 
sin 30 sin 35 sin A sin C
200sin 30 a

6
b
sin 35 sin 65 sin 50
b  174.3 6sin 65
a
The solution is B  30, a  316.0, and b  174.3 . sin 50
a  7.1
14. Begin by finding A.
A  B  C  180 Use the Law of Sines to find angle B.
A  5  125  180 b

c
A  130  180 sin B sin C
b 6
A  50 
sin 65 sin 50
b 200
Use the ratio , or , to find the other two 6sin 65
sin B sin 5 b
sides. Use the Law of Sines to find a. sin 50
a b b  7.1
 The solution is C  50, a  7.1, and b  7.1 .
sin A sin B
a 200

sin 50 sin 5
200sin 50
a
sin 5
a  1757.9

16. Begin by finding A. Use the Law of Sines to find side c.

A  B  C  180 c a

A  80  10  180 sin C sin A
A  90  180 c

20
A  90 sin111 sin 40
20sin111
Use the ratio
a
, or
8
, to find the other two c  29.0
sides. Use the Law of Sines to find b. There is one triangle and the solution is
b a B1 (or B )  29, C  111, and c  29.0 .

sin B sin A
a 30
b 8 18. The known ratio is , or . Use the Law
 sin A sin 50
sin 80 sin 90
of Sines to find angle B.
8sin 80
b a b
sin 90 
sin A sin B
b  7.9
30 20
Use the Law of Sines again, this time to find c. 
c a sin 50 sin B
 30sin B  20sin 50
sin C sin A
c 8 20sin 50
 sin B 
sin10 sin 90 30
8sin10 sin B  0.5107
c There are two angles possible:
sin 90
c  1.4 B1  31, B2  180  31  149
The solution is A  90, b  7.9, and c  1.4 . B2 is impossible, since 50  149  199 .
We find C using B1 and the given information
17. The known ratio is
a
, or
20
. A  50 .
sin A sin 40 C  180  B1  A  180  31  50  99
Use the Law of Sines to find angle B. Use the Law of Sines to find c.
a b c a
 
sin A sin B sin C sin A
20 15 c 30
 
sin 40 sin B sin 99 sin 50
20sin B  15sin 40 30sin 99
c  38.7
15sin 40 sin 50
sin B 
20 There is one triangle and the solution is
sin B  0.4821 B1 (or B )  31, C  99, and c  38.7 .
There are two angles possible:
B1  29, B2  180  29  151
B2 is impossible, since 40  151  191 .
We find C using B1 and the given information A =
40°.
C  180  B1  A  180  29  40  111

a 10 b a
19. The known ratio is , or . 
sin A sin 63 sin B sin A
Use the Law of Sines to find angle C. b 57.5

a

c sin 7 sin136
sin A sin C 57.5sin 7
b  10.1
10

8.9 sin136
sin 63 sin C There is one triangle and the solution is
10sin C  8.9 sin 63 C1 (or C )  37, B  7, and b  10.1 .
8.9 sin 63
sin C  a 42.1
10 21. The known ratio is , or .
sin A sin112
sin C  0.7930 Use the Law of Sines to find angle C.
a c
C1  52, C2  180  52  128 
sin A sin C
C2 is impossible, since 63  128  191 .
42.1 37
We find B using C1 and the given information A = 
sin112 sin C
63°. 42.1sin C  37 sin112
B  180  C1  A  180  52  63  65
37 sin112
Use the Law of Sines to find side b. sin C 
42.1
b a
 sin C  0.8149
sin B sin A There are two angles possible:
b 10 C1  55, C2  180  55  125

sin 65 sin 63 C2 is impossible, since 112  125  237 .
10sin 65
b  10.2 We find B using C1 and the given information A =
sin 63
There is one triangle and the solution is 112°.
C1 (or C )  52, B  65, and b  10.2 . B  180  C1  A  180  55  112  13
Use the Law of Sines to find b.
b a
20. The known ratio is
a
, or
57.5
. 
sin A sin136 sin B sin A
Use the Law of Sines to find angle C. b 42.1

a c sin13 sin112
 42.1sin13
sin A sin C b  10.2
57.5 49.8 sin112
 There is one triangle and the solution is
sin136 sin C
57.5sin C  49.8sin136 C1 (or C )  55, B  13, and b  10.2 .
49.8sin136
sin C 
57.5
sin C  0.6016
C1  37, C2  180  37  143
C2 is impossible, since 136  143  279 .
We find B using C1 and the given information
A  136 .
B  180  C1  A  180  37  136  7
Use the Law of Sines to find b.

a 6.1 a 10
22. The known ratio is , or . 24. The known ratio is , or .
sin A sin162 sin A sin150
Use the Law of Sines to find angle B. Use the Law of Sines to find angle B.
a b a b
 
sin A sin B sin A sin B
6.1 4 10 30
 
sin162 sin B sin150 sin B
6.1sin B  4 sin162 10sin B  30sin150
4 sin162 30sin150
sin B  sin B   1.5
6.1 10
sin B  0.2026 Because the sine can never exceed 1, there is no
There are two angles possible: angle B for which sin B  1.5 . There is no triangle
B1  12, B2  180  12  168 with the given measurements.
B2 is impossible, since 162  168  330 .
a 16
We find C using B1 and the given information 25. The known ratio is , or .
sin A sin 60
A  162 . Use the Law of Sines to find angle B.
C  180  B1  A  180  12  162  6 a b

Use the Law of Sines to find c. sin A sin B
c a 16 18
 
sin C sin A sin 60 sin B
c 6.1 16sin B  18sin 60

sin 6 sin162 18sin 60
6.1sin 6 sin B 
c  2.1 16
sin162 sin B  0.9743
There is one triangle and the solution is There are two angles possible:
B1 (or B )  12, C  6, and c  2.1 . B1  77, B2  180  77  103
There are two triangles:
a 10
23. The known ratio is , or . C1  180  B1  A  180  77  60  43
sin A sin 30
C2  180  B2  A  180  103  60  17 Use the
Use the Law of Sines to find angle B.
a b Law of Sines to find c1 and c2 .

sin A sin B c1 a

10 40 sin C1 sin A

sin 30 sin B c1 16

10sin B  40sin 30 sin 43 sin 60
40sin 30 16sin 43
sin B  2 c1   12.6
10 sin 60
Because the sine can never exceed 1, there is no c2 a
angle B for which sin B = 2. There is no triangle with 
sin C2 sin A
the given measurements.
c2 16

sin17 sin 60
16sin17
c2   5.4
sin 60
In one triangle, the solution is
B1  77, C1  43, and c1  12.6 .
In the other triangle,
B2  103, C2  17, and c2  5.4 .

a 30 C1  180  B1  A  180  54  37  89

26. The known ratio is , or .
sin A sin 20 C2  180  B2  A  180  126  37  17 Use the
Use the Law of Sines to find angle B. Law of Sines to find c1 and c2 .
a b
 c1 a
sin A sin B 
sin C1 sin A
30 40
 c1 12
sin 20 sin B 
30sin B  40sin 20 sin 89 sin 37
12 sin 89
40sin 20 c1   19.9
sin B  sin 37
30
c2 a
sin B  0.4560 
There are two angles possible: sin C2 sin A
B1  27, B2  180  27  153 c2 12

There are two triangles: sin17 sin 37
C1  180  B1  A  180  27  20  133 12 sin17
c2   5.8
C2  180  B2  A  180  153  20  7 sin 37
Use the Law of Sines to find c1 and c2 . In one triangle, the solution is
B1  54, C1  89, and c1  19.9 .
c1 a
 In the other triangle,
sin C1 sin A
B2  126, C2  17, and c2  5.8 .
c1 30

sin133 sin 20 a 7
30sin133 28. The known ratio is , or .
c1   64.2 sin A sin12
sin 20 Use the Law of Sines to find angle B.
c2 a a b
 
sin C2 sin A sin A sin B
c2 30 7 28
 
sin 7 sin 20 sin12 sin B
30sin 7 7 sin B  28sin12
c2   10.7
sin 20 28sin12
sin B 
In one triangle, the solution is 7
B1  27, C1  133, and c1  64.2 . sin B  0.8316
In the other triangle, There are two angles possible:
B2  153, C2  7, and c2  10.7 . B1  56, B2  180  56  124
a 12 C1  180  B1  A  180  56  12  112
27. The known ratio is , or .
sin A sin 37 C2  180  B2  A  180  124  12  44
Use the Law of Sines to find angle B. Use the Law of Sines to find c1 and c2 .
a b

sin A sin B
12 16.1

sin 37 sin B
12 sin B  16.1sin 37
16.1sin 37
sin B 
12
sin B  0.8074
B1  54, B2  180  54  126

c1 a In one triangle, the solution is

 C1  68, B1  54, and b1  21.0 .
sin C1 sin A
c1 7 In the other triangle,
 C2  112, B2  10, and b2  4.5 .
sin112 sin12
7 sin112
c1   31.2 a 95
sin12 30. The known ratio is , or .
sin A sin 49
c2 a
 Use the Law of Sines to find angle C.
sin C2 sin A a c
c2 7 
 sin A sin C
sin 44 sin12 95 125
7 sin 44 
c2   23.4 sin 49 sin C
sin12 95sin C  125sin 49
In one triangle, the solution is 125sin 49
B1  56, C1  112, and c1  31.2 . sin C 
95
In the other triangle,
sin C  0.9930
B2  124, C2  44, and c2  23.4 . There are two angles possible:
C1  83, C2  180  83  97
a 22
29. The known ratio is , or . There are two triangles:
sin A sin 58 B1  180  C1  A  180  83  49  48
Use the Law of Sines to find angle C.
B2  180  C2  A  180  97  49  34
a c
 Use the Law of Sines to find b1 and b2 .
sin A sin C
22 24.1 b1 a
 
sin 58 sin C sin B1 sin A
22 sin C  24.1sin 58 b1 95

24.1sin 58 sin 48 sin 49
sin C 
22 95sin 48
b1   93.5
sin C  0.9290 sin 49
There are two angles possible: b2 a

C1  68, C2  180  68  112 sin B2 sin A
There are two triangles: b2 95
B1  180  C1  A  180  68  58  54 
sin 34 sin 49
B2  180  C2  A  180  112  58  10 Use the 95sin 34
b2   70.4
Law of Sines to find b1 and b2 . sin 49
b1 a In one triangle, the solution is
 C1  83, B1  48, and b1  93.5 .
sin B1 sin A
b1 In the other triangle,
22
 C2  97, B2  34, and b2  70.4 .
sin 54 sin 58
22 sin 54
b1   21.0
sin 58
b2 a

sin B2 sin A
b2 22

sin10 sin 58
22 sin10
b2   4.5
sin 58

a 9.3 1 1
31. The known ratio is , or . 38. Area  ab sin C  (16)(20)(sin102)  157
sin A sin18 2 2
Use the Law of Sines to find angle B. The area of the triangle is approximately 157 square
a b meters.

sin A sin B
39. ABC  180  67  113
9.3 41
 ACB  180  43  113  24
sin18 sin B
Use the law of sines to find BC .
9.3sin B  41sin18
BC 312
41sin18 
sin B   1.36 sin 43 sin 24
9.3
312sin 43
Because the sine can never exceed 1, there is no BC 
angle B for which sin B = 1.36. There is no triangle sin 24
with the given measurements. BC  523.1
Use the law of sines to find h.
a 1.4 h 523.1
32. The known ratio is , or . 
sin A sin142 sin 67 sin 90
Use the Law of Sines to find angle B. 523.1sin 67
a b h
 sin 90
sin A sin B h  481.6
1.4 2.9

sin142 sin B 40. ABC  180  29  151
1.4sin B  2.9sin142 ACB  180  25  151  4
2.9sin142 Use the law of sines to find BC .
sin B   1.28
1.4 BC 238

Because the sine can never exceed 1, there is no sin 25 sin 4
angle B for which sin B  1.28 . There is no triangle 238sin 25
with the given measurements. BC 
sin 4
1 1 BC  1441.9
33. Area  bc sin A  (20)(40)(sin 48)  297 Use the law of sines to find h.
2 2
h 1441.9
The area of the triangle is approximately 
297 square feet. sin 29 sin 90
1441.9sin 29
h
34. Area 
1 1
bc sin A  (20)(50)(sin 22)  187 sin 90
2 2 h  699.1
The area of the triangle is approximately 187 square
feet. 41. Begin by finding the six angles inside the two
triangles. Then use the law of sines.
1 1 450sin145
35. Area  ac sin B  (3)(6)(sin 36)  5
 sin 34
2 2 a
The area of the triangle is approximately sin 4 sin 30
5 square yards. a  64.4
1 1 42. Begin by finding the six angles inside the two
36. Area  ac sin B  (8)(5)(sin125)  16
2 2 triangles. Then use the law of sines.
The area of the triangle is approximately 16 square 120
yards. a
 sin 58
sin 22 sin100
1 1 a  53.8
37. Area  ab sin C  (4)(6)(sin124)  10
2 2
The area of the triangle is approximately
10 square meters.

a b 46.
43. 
sin A sin B
300 200

sin 2θ sin θ
200sin 2θ  300sin θ
400sin θ cos θ  300sin θ
300sin θ
cos θ 
400sin θ
3
cos θ  1
4 A bh
2
θ  41
1
2θ  82  (5)(4)
2
A  82, B  41, C  57, c  255.7
 10
a b
44.  47.
sin A sin B
400 300

sin 2θ sin θ
300sin 2θ  400sin θ
600sin θ cos θ  400sin θ
400sin θ
cos θ 
600sin θ
2
cos θ 
3 Using a north-south line, the interior angles are found
θ  48 as follows:
2θ  96 A  90  25  65
A  96, B  48, C  36, c  237.3 B  90  56  34
Find angle C using a 180° angle sum in the triangle.
45. C  180  A  B  180  65  34  81
c 10
The ratio , or , is now known. Use this
sin C sin 81
ratio and the Law of Sines to find b and a.
b c

sin B sin C
b 10

sin 34 sin 81
10sin 34
1 b  5.7
A bh sin 81
2 Station A is about 5.7 miles from the fire.
1
 (5)(4) a

c
2 sin A sin C
 10 a 10

sin 65 sin 81
10sin 65
a  9.2
sin 81
Station B is about 9.2 miles from the fire.

48. 49.
Using a north-south line, the interior angles are found

as follows: Using the figure,
A  90  48  42 C  180  A  B  180  85  76  19
B  90  34  56 c 1200
The ratio , or , is now known. Use this
Find angle C using a 180º angle sum in the triangle. sin C sin19
C  180  A  B  180  42  56  82 ratio and the Law of Sines to find a and b.
c 40 a c
The ratio , or , is now known. 
sin C sin 82 sin A sin C
Use this ratio and the Law of Sines to find a 1200
b and a. 
sin 85 sin19
b c
 1200sin 85
sin B sin C a  3671.8
sin19
b 40
 b

c
sin 56 sin 82 sin B sin C
40sin 56
b  33.5 b

1200
sin 82 sin 76 sin19
Station A is about 33.5 miles from the illegal station.
1200sin 76
a c b  3576.4
 sin19
sin A sin C
The platform is about 3671.8 yards from one end of
a 40
 the beach and 3576.4 yards from the other.
sin 42 sin 82
40sin 42 50. Let c = distance from A to B .
a  27.0 Using the figure,
sin 82
Station B is about 27.0 miles from the illegal station. B  180  A  C  180  62  53  65
b 300
The ratio , or , is now known.
sin B sin 65
Use this ratio and the Law of Sines to find c.
c b

sin C sin B
c 300

sin 53 sin 65
300sin 53
c  264.4
sin 65
The distance between A and B is about 264.4 yards or
793 feet.

51. According to the figure, 54. Using the figure,

C  180  A  B  180  84.7  50  45.3 The B  85  180
c 171 B  95
ratio , or , is now known. Use this
sin C sin 45.3 A  B  C  180
ratio and the Law of Sines to find b.
37  95  C  180
b c
 132  C  180
sin B sin C
b 171 C  48
 c 100
sin 50 sin 45.3 The ratio , or , is now known.
171sin 50 sin C sin 48
b  184 Use this ratio and the Law of Sines to find a.
sin 45.3
a c
The distance is about 184 feet. 
sin A sin C
52. a 100

sin 37 sin 48
100sin 37
a  81.0
sin 48
The pier is about 81.0 feet long.
55.
Using the figure,

C  62  23  85
B  180  A  C  180  75  85  20
b 80
The ratio , or , is now known.
sin B sin 20
Use this ratio and the Law of Sines to find c. Using the figure,
c b B  90  8  82

sin C sin B C  180  A  B  180  62  82  36
c 80 c 20
 The ratio , or , is now known. Use this
sin 85 sin 20 sin C sin 36
80sin 85 ratio and the Law of Sines to find a.
c  233.0 a c
sin 20 
The height of the tree is about 233.0 feet. sin A sin C
a 20

53. The ratio
b
, or
562
, is known. sin 62 sin 36
sin B sin 85.3 20sin 62
Use this ratio, the figure, and the Law of Sines to find a  30.0
sin 36
c. The length of the pole is about 30.0 feet.
c b

sin C sin B
c 562

sin 5.7 sin 85.3
562sin 5.7
c  56.0
sin 85.3
The toss was about 56.0 feet.

56. Using the figure, 58. a. Using the figure,

A  90  6  84 B  180  66  114
C  180  A  B  180  84  22  74 C  180  A  B  180  22  114  44
c 40 c 1.6
The ratio , or , is now known. The ratio , or , is now known.
sin C sin 74 sin C sin 44
Use this ratio and the Law of Sines to find b. Use this ratio and the Law of Sines to find b.
b c b c
 
sin B sin C sin B sin C
b 40 b 1.6
 
sin 22 sin 74 sin114 sin 44
40sin 22 1.6sin114
b  15.6 b  2.1042
sin 74 sin 44
The height of the wall is about 15.6 feet. The cable car covers about 2.1042 miles.
There are 5280 feet per mile, so the cable car
57. a. Using the figure and the measurements shown, covers about 11,110.2 feet.
B  180  44  136
C  180  B  A  180  136  37  7 b. The known ratio is
c
, or
1.6
.
sin C sin 44
c 100 Use the Law of Sines to find a.
The ratio , or , is now known. Use
sin C sin 7 a c
this ratio and the Law of Sines to find a. 
sin A sin C
a c
 a 1.6
sin A sin C 
sin 22 sin 44
a 100
 1.6sin 22
sin 37 sin 7 a  0.8628
sin 44
100sin 37
a  493.8 0.8628 miles ≈ 4555.6 feet
sin 7 a ≈ 4555.6 feet
To the nearest foot, a = 493.8 feet.
c. Let a = 4555.6, to the nearest foot, be the
b. From part a, let a = 493.8 be the hypotenuse of
the right triangle. hypotenuse of the right triangle. Then if h
If h represents the height of the tree, represents the height of the mountain,
h a
h

493.8 
sin 44 sin 90 sin 66 sin 90
493.8sin 44 h 4555.6
h  343.0 
sin 90 sin 66 sin 90
A typical redwood tree is about 343.0 feet. 4555.6sin 66
h  4161.7
sin 90
The mountain is about 4161.7 feet high.

59. 60.
a
Using the figure, the known ratio is , or
sin A
Using the figure,
B  90  62  28 16
. Use this ratio and the Law of Sines to find
b 5 sin 48
The known ratio is , or . C.
sin B sin 28
a c
Use the Law of Sines to find angle C. 
b c sin A sin C
 16 15
sin B sin C 
5 7 sin 48 sin C
 16sin C  15sin 48
sin 28 sin C
5sin C  7sin 28 15sin 48
sin C   0.6967
7sin 28 16
sin C   0.6573
5 There are two angles possible:
C1  44, C2  180  44  136
C1  41, C2  180  41  139
C2 is impossible, since 48  136  184
A1  180  C1  B  180  41  28  111 B  180  48  44  88
Use the The flagpole is leaning because it makes about an 88
A2  180  C2  B  180  139  28  13
angle with the ground.
Law of Sines to find a1 and a2 .
a1 b 61. – 70. Answers may vary.

sin A1 sin B 71. does not make sense; Explanations will vary.
a1 5 Sample explanation: The law of cosines would be
 appropriate for this situation.
sin111 sin 28
5sin111
a1   9.9 72. makes sense
sin 28
a2 b 73. does not make sense; Explanations will vary.
 Sample explanation: The calculator will give you
sin A2 sin B
the acute angle. The obtuse angle is the supplement
a2 5
 of the acute angle.
sin13 sin 28
5sin13 74. makes sense
a2   2.4
sin 28
75. No. Explanations may vary.
The boat is either 9.9 miles or 2.4 miles from
lighthouse B, to the nearest tenth of a mile.

76.
Let h = the height of the buildings. Using the figure, b  e  800

e  800  b
Now the law of sines gives the following equations:
b h
 (1)
sin 63 sin 27
800  b h
 (2)
sin 49 sin 41
Solve (1) for b:
b h

sin 63 sin 27
h sin 63
b
sin 27
Now substitute into (2):
800  b h

sin 49 sin 41
h sin 63
800 
sin 27  h
sin 49 sin 41
800sin 27  h sin 63 h

sin 27 sin 49 sin 41
h sin 27 sin 49  sin 41(800sin 27)  h sin 63 sin 41
h sin 27 sin 49  h sin 63 sin 41  sin 41(800sin 27)
h(sin 27 sin 49  sin 63 sin 41)  800sin 41 sin 27
800sin 41 sin 27
h  257
sin 27 sin 49  sin 63 sin 41
The buildings are about 257 feet high.
77.
Using the figure, A  180  150  30

Using the Law of Sines we have,
d 36

sin A sin 90
d 36

sin 30 sin 90
36sin 30
d  18
sin 90
CC   18  5  18  41
The wingspan CC  is 41 feet.

62  4 2  9 2 b

a
78. cos B 
264 sin B sin A
29 7

13
cos B 
48 sin B sin120
29 13sin B  7sin120
cos B 
48 7sin120
sin B   0.4663
 29  13
B  cos 1 
 48  B  28
Find the third angle.
B  127 C  180  A  B  180  120  28  32
The solution is a  13, B  28, and C  32 .
79. 26(26  12)(26  16)(26  24)
 26(14)(10)(2) 2. Apply the three-step procedure for solving a SSS
triangle. Use the Law of Cosines to find the angle
 7280 opposite the longest side.
 4 455 Thus, we will find angle B.
b2  a 2  c 2  2ac cos B
 85
2ac cos B  a 2  c 2  b2
80. Diagram:
a 2  c 2  b2
cos B 
2ac
8  52  102
2
11
cos B  
285 80
 11 
cos1    82
 80 
B is obtuse, since cos B is negative.
B  180  82  98
Section 7.2 Use the Law of Sines to find either of the two
remaining acute angles. We will find angle A.
Check Point Exercises a b

1. Apply the three-step procedure for solving sin A sin B
a SAS triangle. Use the Law of Cosines to find the 8 10

side opposite the given angle. sin A sin 98
Thus, we will find a. 10sin A  8sin 98
a 2  b2  c 2  2bc cos A 8sin 98
sin A   0.7922
a  7  8  2(7)(8) cos120
2 2 2 10
 49  64  112( 0.5) A  52
 169 C  180  A  B  180  52  98
a  169  13  30
Use the Law of Sines to find the angle opposite the The solution is B  98, A  52, and C  30
shorter of the two sides. Thus, we will find acute
angle B.

Section 7.2 The Law of Cosines
3. The plane flying 400 miles per hour travels Exercise Set 7.2
400  2  800 miles in 2 hours. Similarly, the other
plane travels 700 miles. 1. Apply the three-step procedure for solving a SAS triangle.
Use the Law of Cosines to find the side opposite the given
angle.
Thus, we will find a.
a 2  b2  c 2  2bc cos A
a 2  42  82  2(4)(8) cos 46
a 2  16  64  64(cos 46)
a 2  35.54
a  35.54  6.0
Use the Law of Sines to find the angle opposite the shorter
Use the figure and the Law of Cosines to find a in of the two given sides. Thus, we will find acute angle B.
this SAS situation. b a

a 2  b2  c 2  2bc cos A sin B sin A
a 2  7002  8002  2(700)(800) cos 75 4 35.54

 840,123 sin B sin 46
a  840,123  917 35.54 sin B  4 sin 46
After 2 hours, the planes are approximately 917 miles 4 sin 46
sin B   0.4827
apart. 35.54
B  29
4. Begin by calculating one-half the perimeter:
1 1 C  180  A  B  180  46  29  105
s  ( a  b  c)  (6  16  18)  20
2 2 The solution is a  6.0, B  29, and C  105 .
Use Heron’s formula to find the area.
Area  s ( s  a )( s  b)( s  c ) 2. Apply the three-step procedure for solving a SAS triangle.
Use the Law of Cosines to find the side opposite the given
 20(20  6)(20  16)(20  18) angle. Thus, we will find b.
 2240  47 b2  a 2  c 2  2ac cos B
b2  62  82  2(6)(8) cos 32
47 square meters.
b2  36  64  96 cos 32
b2  18.59
Concept and Vocabulary Check 7.2 b  18.59  4.3
Use the Law of Sines to find the angle opposite the shorter
1. b2  c 2  2b cos A of the two given sides. Thus, we will find acute angle A.
a b
2. side; Cosines; Sines; acute; 180° 
sin A sin B
3. Cosines; Sines 6 18.59

sin A sin 32
1
4. s ( s  a )( s  b)( s  c ) ; (a  b  c) 18.59 sin A  6sin 32
2
6sin 32
sin A   0.7374
18.59
A  48
C  180  A  B  180  48  32  100
The solution is b  4.3, A  48, and C  100 .

3. Apply the three-step procedure for solving a SAS 5. Apply the three-step procedure for solving a SSS
triangle. Use the Law of Cosines to find the side triangle. Use the Law of Cosines to find the angle
opposite the given angle. opposite the longest side. Since two sides have length
Thus, we will find c. 8, we can begin by finding angle B or C.
c 2  a 2  b2  2ab cos C b2  a 2  c 2  2ac cos B
c 2  62  42  2(6)(4) cos 96 a 2  c 2  b2
cos B 
c 2  36  16  48(cos 96) 2ac
6  82  82 36 3
2
c 2  57.02 cos B   
268 96 8
c  57.02  7.6 B  68
Use the Law of Sines to find the angle opposite the Use the Law of Sines to find either of the two
shorter of the two given sides. Thus, we will find remaining acute angles. We will find angle A.
acute angle B.
a b
b c 
 sin A sin B
sin B sin C
6 8
4 57.02 
 sin A sin 68
sin B sin 96 8sin A  6sin 68
57.02 sin B  4 sin 96 6sin 68
4 sin 96 sin A   0.6954
sin B   0.5268 8
57.02 A  44
B  32 Find the third angle.
Find the third angle. C  180  B  A  180  68  44  68
A  180  B  C  180  32  96  52 The solution is A  44, B  68, and C  68 .
The solution is c  7.6, A  52, and B  32 .
6. Apply the three-step procedure for solving a SSS
4. Apply the three-step procedure for solving a SAS
opposite the longest side. Thus, we will find C.
triangle. Use the Law of Cosines to find the side
opposite the given angle. Thus, we will find a. c 2  a 2  b2  2ab cos C
a 2  b2  c 2  2bc cos A a 2  b2  c 2
cos C 
a 2  62  152  2(6)(15) cos 22 2ab
10  122  162
2
12
a 2  36  225  180(cos 22) cos C  
2  10  12 240
a 2  94.11 C is obtuse, since cos C is negative.
a  94.11  9.7  12 
cos1   87
Use the Law of Sines to find the angle opposite the  240 
shorter of the two given sides. Thus, we will find
C  180  87  93
acute angle B.
Use the Law of Sines to find either of the two
b a
 remaining acute angles. We will find angle B.
sin B sin A b c

6 94.11 sin B sin C

sin B sin 22 12 16

94.11 sin B  6sin 22 sin B sin 93
6sin 22 16sin B  12 sin 93
sin B   0.2317
94.11 12 sin 93
sin B   0.7490
B  13 16
Find the third angle. B  49
C  180  A  B  180  22  13  145 Find the third angle.
The solution is a  9.7, B  13, and C  145 . A  180  B  C  180  49  93  38
The solution is A  38, B  49, and C  93 .

7. Apply the three-step procedure for solving a SSS Use the Law of Sines to find either of the two
triangle. Use the Law of Cosines to find the angle remaining acute angles. We will find angle A.
opposite the longest side. Thus, we will find angle A a b

a 2  b2  c 2  2bc cos A sin A sin B
b2  c 2  a 2 10 16
cos A  
2bc sin A sin125
16sin A  10sin125
4  32  62
2
11
cos A   10sin125
243 24 sin A   0.5120
A is obtuse, since cos A is negative. 16
 11  A  31
cos1    63 Find the third angle.
 24 
C  180  B  A  180  125  31  24
A  180  63  117 The solution is B  125, A  31, and C  24 .
remaining acute angles. We will find angle B. 9. Apply the three-step procedure for solving a SAS
b a triangle. Use the Law of Cosines to find the side

sin B sin A opposite the given angle.
4 6 Thus, we will find c.

sin B sin117 c 2  a 2  b2  2ab cos C
6sin B  4 sin117 c 2  52  72  2(5)(7) cos 42
4 sin117 c 2  25  49  70(cos 42)
sin B   0.5940
6
c 2  21.98
B  36
Find the third angle. c  21.98  4.7
C  180  B  A  180  36  117  27 Use the Law of Sines to find the angle opposite the
The solution is A  117, B  36, and C  27 . shorter of the two given sides. Thus, we will find
acute angle A.
a c
triangle. Use the Law of Cosines to find the angle 
opposite the longest side. Thus, we will find B. sin A sin C
b2  a 2  c 2  2ac cos B 5 4.7

sin A sin 42
a 2  c2  b2 4.7 sin A  5sin 42
cos B 
2ac
5sin 42
10  82  162
2
23 sin A   0.7118
cos B   4.7
2  10  8 40 A  45
B is obtuse, since cos B is negative. Find the third angle.
 23  B  180  C  A  180  42  45  93
cos1    55
 40  The solution is c  4.7, A  45,and B  93 .
B  180  55  125

10. Apply the three-step procedure for solving a SAS 12. Apply the three-step procedure for solving a SAS
triangle. Use the Law of Cosines to find the side triangle. Use the Law of Cosines to find the side
opposite the given angle. Thus, we will find c. opposite the given angle. Thus, we will find a.
c 2  a 2  b2  2ab cos C a 2  b2  c 2  2bc cos A
c 2  102  32  2(10)(3) cos15 a 2  42  12  2(4)(1) cos100
c 2  100  9  60(cos15) a 2  16  1  8(cos100)
c 2  51.04 a 2  18.39
c  51.04  7.1 a  18.39  4.3
Use the Law of Sines to find the angle opposite the Use the Law of Sines to find the angle opposite the
shorter of the two given sides. Thus, we will find shorter of the two given sides. Thus, we will find
acute angle B. acute angle C.
b c c a
 
sin B sin C sin C sin A
3 7.1 1 4.3
 
sin B sin15 sin C sin100
7.1sin B  3sin15 4.3sin C  sin100
3sin15 sin100
sin B   0.1094 sin C   0.2290
7.1 4.3
B  6 C  13
Find the third angle. Find the third angle.
A  180  C  B  180  15  6  159 B  180  C  A  180  13  100  67
The solution is c  7.1, B  6, and A  159 . The solution is a  4.3, C  13, and B  67 .
triangle. Use the Law of Cosines to find the side triangle. Use the Law of Cosines to find the side
opposite the given angle. opposite the given angle.
Thus, we will find a. Thus, we will find b.
a 2  b2  c 2  2bc cos A b2  a 2  c 2  2ac cos B
a 2  52  32  2(5)(3) cos102 b2  62  52  2(6)(5) cos 50
a 2  25  9  30(cos102) b2  36  25  60(cos50)
a 2  40.24 b2  22.43
a  40.24  6.3 b  22.43  4.7
Use the Law of Sines to find the angle opposite the Use the Law of Sines to find the angle opposite the
shorter of the two given sides. Thus, we will find shorter of the two given sides. Thus, we will find
acute angle C. acute angle C.
c a c b
 
sin C sin A sin C sin B
3 6.3 5 4.7
 
sin C sin102 sin C sin 50
6.3sin C  3sin102 4.7 sin C  5sin 50
3sin102 5sin 50
sin C   0.4658 sin C   0.8149
6.3 4.7
C  28 C  55
B  180  C  A  180  28  102  50 A  180  C  B  180  55  50  75
The solution is a  6.3, C  28, and B  50 . The solution is b  4.7, C  55, and A  75 .

triangle. Use the Law of Cosines to find the side triangle. Use the Law of Cosines to find the side opposite
opposite the given angle. Thus, we will find b. the given angle. Thus, we will find b.
b2  a 2  c 2  2ac cos B b2  a 2  c 2  2ac cos B
b2  42  72  2(4)(7) cos55 b2  72  32  2(7)(3) cos 90
b2  16  49  56(cos 55) b2  49  9  42 cos 90
b2  32.88 b2  58
b  32.88  5.7 b  58  7.6
Use the Law of Sines to find the angle opposite the (use exact value of b from previous step)
shorter of the two given sides. Thus, we will find Use the Law of Sines to find the angle opposite the shorter
acute angle A. of the two given sides. Thus, we will find acute angle C.
a b c b
 
sin A sin B sin C sin B
4 5.7 3 7.6
 
sin A sin 55 sin C sin 90
5.7 sin A  4 sin 55 7.6sin C  3sin 90
4 sin 55 3sin 90
sin A   0.5749 sin C   0.3947
5.7 7.6
A  35 C  23
C  180  B  A  180  55  35  90 A  180  C  B  180  23  90  67
The solution is b  5.7, A  35, and C  90 . The solution is b  7.6, C  23, and A  67 .
15. Apply the three-step procedure for solving a SAS 17. Apply the three-step procedure for solving a SSS
triangle. Use the Law of Cosines to find the side triangle. Use the Law of Cosines to find the angle
opposite the given angle. opposite the longest side. Thus, we will find C.
Thus, we will find b. c 2  a 2  b2  2ab cos C
b2  a 2  c 2  2ac cos 90 a 2  b2  c 2
cos C 
b2  52  22  2(5)(2) cos 90 2ab
b2  25  4  20 cos 90 5  72  102
2
13
cos C  
2  5 7 35
b2  29 C is obtuse, since cos C is negative.
b  29  5.4  13 
cos1    68
(use exact value of b from previous step) Use the  35 
Law of Sines to find the angle opposite the shorter of
the two given sides. Thus, we will find acute angle C. C  180  68  112
c b
 remaining angles. We will find angle A.
sin C sin B a c
2 5.4 
 sin A sin C
sin C sin 90 5 10
5.4 sin C  2 sin 90 
sin A sin112
2 sin 90 10sin A  5sin112
sin C   0.3704
5.4 5sin112
C  22 sin A   0.4636
10
A  180  C  B  180  22  90  68 A  28
The solution is b  5.4, C  22, and A  68 .
B  180  C  A  180  112  28  40
The solution is C  112, A  28, and B  40 .

18. Apply the three-step procedure for solving a SSS Find the third angle.
triangle. Use the Law of Cosines to find the angle C  180  B  A  180  100  19  61
opposite the longest side. Thus, we will find C. The solution is B  100, A  19, and C  61 .
c 2  a 2  b2  2ab cos C
a 2  b2  c 2
cos C  triangle. Use the Law of Cosines to find the angle
2ab opposite the longest side. Thus, we will find B.
4  62  9 2
2
29 b2  a 2  c 2  2ac cos B
cos C  
246 48 a 2  c2  b2
C is obtuse, since cos C is negative. cos B 
2ac
 29 
cos1    53 4 2  62  7 2 1
 48  cos B  
246 16
C  180  53  127
B  86
remaining angles. We will find angle A.
remaining angles. We will find angle A.
a c
 a b
sin A sin C 
sin A sin B
4 9
 4 7
sin A sin127 
sin A sin 86
9 sin A  4 sin127
7 sin A  4 sin 86
4 sin127
sin A   0.3549 4 sin 86
9 sin A   0.5700
7
A  21
A  35
B  180  C  A  180  127  21  32
C  180  B  A  180  86  35  59
The solution is C  127, A  21, and B  32 .
The solution is B  86, A  35, and C  59 .
triangle. Use the Law of Cosines to find any of the
opposite the longest side. Thus, we will find B.
three angles, since each side has the same measure.
b2  a 2  c 2  2ac cos B
a 2  b2  c 2  2bc cos A
a 2  c2  b2
cos B  b2  c2  a 2
2ac cos A 
2bc
3  82  9 2
2
1
cos B   32  32  32 1
2  3 8 6 cos A  
2  3 3 2
B is obtuse, since cos B is negative.
A  60
1
cos1    80 Use the Law of Sines to find either of the two
6 remaining angles. We will find angle B.
B  180  80  100 b a

Use the Law of Sines to find either of the two sin B sin A
remaining angles. We will find angle A. 3 3
a b 
 sin B sin 60
sin A sin B 3sin B  3sin 60
3 9
 sin B  sin 60
sin A sin100
B  60
9 sin A  3sin100
3sin100 C  180  A  B  180  60  60  60
sin A   0.3283
9 The solution is A  60, B  60, and C  60 .
A  19

22. Apply the three-step procedure for solving a SSS 24. Apply the three-step procedure for solving a SSS
triangle. Use the Law of Cosines to find any of the triangle. Use the Law of Cosines to find the angle
three angles, since each side has the same measure. opposite the longest side. Thus, we will find A.
a 2  b2  c 2  2bc cos A a 2  b2  c 2  2bc cos A
b2  c2  a 2 b2  c2  a 2
cos A  cos A 
2bc 2bc
5  52  52 1
2
25  452  662
2
853
cos A   cos A  
2  5 5 2 2  25  45 1125
A  60 A is obtuse, since cos A is negative.
Use the Law of Sines to find either of the two  853 
cos1   41
remaining angles. We will find angle B.  1125 
b a
 A  180  41  139
sin B sin A Use the Law of Sines to find either of the two
5 5 remaining angles. We will find angle B.

sin B sin 60 b a

5sin B  5sin 60 sin B sin A
sin B  sin 60 25 66

B  60 sin B sin139
Find the third angle. 66sin B  25sin139
C  180  A  B  180  60  60  60 25sin139
The solution is A  60, B  60, and C  60 . sin B   0.2485
66
B  14
C  180  A  B  180  139  14  27
opposite the longest side. Thus, we will find A.
The solution is A  139, B  14, and C  27 .
a 2  b2  c 2  2bc cos A
b2  c 2  a 2 1 1
cos A  25. s ( a  b  c)  (4  4  2)  5
2bc 2 2
22  502  632
2
985 Area  s ( s  a )( s  b)( s  c )
cos A  
2  22  50 2200  5(5  4)(5  4)(5  2)
A  117
Use the Law of Sines to find either of the two  15  4
remaining angles. We will find angle B. The area of the triangle is approximately
b a 4 square feet.

sin B sin A 1 1
22 63 26. s ( a  b  c )  (5  5  4)  7
 2 2
sin B sin117 Area  s ( s  a )( s  b)( s  c )
63sin B  22 sin117
 7(7  5)(7  5)(7  4)
22 sin117
sin B 
63  84  9
B  18 The area of the triangle is approximately
Find the third angle. 9 square feet.
C  180  A  B  180  117  18  45
The solution is A  117, B  18, and C  45 .

1 1 33. Use the given radii to determine that

27. s ( a  b  c)  (14  12  4)  15 a  BC  7.5, b  AC  8.5, and c  AB  9.0 .
2 2
Area  s ( s  a )( s  b)( s  c ) c 2  a 2  b2  2ab cos C
 15(15  14)(15  12)(15  4) 92  7.52  8.52  2(7.5)(8.5) cos C
 495  22 cos C  0.3725
The area of the triangle is approximately C  68
22 square meters. Use the law of sines to find the solution is
A  51, B  61, and C  68.
1 1
28. s ( a  b  c)  (16  10  8)  17
2 2 34. Use the given radii to determine that
Area  s ( s  a )( s  b)( s  c ) a  BC  7.3, b  AC  10.5, and c  AB  11.8 .
 17(17  16)(17  10)(17  8) c 2  a 2  b2  2ab cos C
 1071  33 11.82  7.32  10.52  2(7.3)(10.5) cos C

The area of the triangle is approximately cos C  0.1585
33 square meters. C  81
Use the law of sines to find the solution is
1 1
29. s ( a  b  c )  (11  9  7)  13.5 A  38, B  61, and C  81.
2 2
Area  s ( s  a )( s  b)( s  c ) 35. Use the distance formula to determine that
 13.5(13.5  11)(13.5  9)(13.5  7) a  61  7.8, b  10  3.2, and c  5 .
 987.1875  31 a 2  b2  c 2  2bc cos A

31 square yards.
2 2
61  10  52  2  10  (5) cos A
cos A  0.8222
1 1
30. s  ( a  b  c )  (13  9  5)  13.5 A  145
2 2 Use the law of sines to find the solution is
Area  s ( s  a )( s  b)( s  c ) A  145, B  13, and C  22.
 13.5(13.5  13)(13.5  9)(13.5  5)
36. Use the distance formula to determine that
 258.1875  16 a  13  3.6, b  26  5.1, and c  5 .
16 square yards. b2  a 2  c 2  2ac cos B
31. C  180  15  35  130

2 2
26  13  52  2  13  (5) cos B
c 2  b2  c 2  2bc cos C cos B  0.3328
c  8  13  2(8)(13) cos130
2 2 2 B  71
c 2  366.6998 A  42, B  71, and C  67.
c  19.1
Use the law of sines to find the solution is 37. Use the law of cosines.
A  31, B  19, C  130, and c  19.1. c 2  a 2  b2  2ab cos C
32. C  180  35  50  95 5.782  2.92  3.02  2(2.9)(3.0) cos θ
c 2  a 2  b2  2ab cos C cos θ  0.9194
c  3  2  2(3)(2) cos 95
2 2 2 θ  157
This dinosaur was an efficient walker.
c 2  14.0459
c  3.7
A  54, B  31, C  95, and c  3.7.

38. Use the law of cosines. 41. Let b = the distance across the lake.
c 2  a 2  b2  2ab cos C b2  a 2  c 2  2ac cos B
5.22  3.62  3.22  2(3.6)(3.2) cos θ b2  1602  1402  2(160)(140) cos80
cos θ  0.1667  37, 421
θ  100 b  37, 421  193
This dinosaur was not an efficient walker.
The distance across the lake is about
39. Let b = the distance between the ships after three hours. 193 yards.
After three hours, the ship traveling
14 miles per hour has gone 3  14 or 42. Let c = the distance from A to B.
42 miles. Similarly, the ship traveling c 2  a 2  b2  2ab cos C
10 miles per hour has gone 30 miles. c 2  1052  652  2(105)(65) cos80  12,880
c  12,880  113
The distance from A to B is about 113 yards.
43. Assume that Island B is due east of Island A. Let

A = angle at Island A.
a 2  b2  c 2  2bc cos A
b2  c2  a 2
cos A 
2bc
52  62  72 1
cos A  
2  5 6 5
Using the figure,
A  78
B  180  75  12  117
Since 90  78  12, you should navigate on a
b2  a 2  c 2  2ac cos B bearing of N12°E.
b2  302  422  2(30)(42) cos117  3808
44. Assume that Island A is due west of Island B. Let
b  61.7 B = angle at Island B.
After three hours, the ships will be about
b2  a 2  c 2  2ac cos B
61.7 miles apart.
a 2  c2  b2
40. First, make a diagram. cos B 
2ac
72  62  52 5
cos B  
276 7
B  44
Since 90  44  46, you should navigate on a
bearing of N46°W.
45. a. Using the figure,

B  90  40  50
b2  a 2  c 2  2ac cos B
Using the diagram, b2  13.52  252  2(13.5)(25) cos50
B  74  34  108
 373
b2  a 2  c 2  2ac cos B
b  373  19.3
b2  4002  5802  2(400)(580) cos108 You are about 19.3 miles from the pier.
 639, 784
b  639,784  799.9
The distance from airport A to airport B is about
799.9 miles.

a b In the figure, b = the guy wire anchored downhill,

b.  e = the guy wire anchored uphill.
sin A sin B
B  90  7  97
13.5 373
 E  90  7  83
sin A sin 50
373 sin A  13.5sin 50 b2  a 2  c 2  2ac cos B
13.5sin 50 b2  4002  802  2(400)(80) cos 97
sin A   0.5355
373  174, 200
A  32 b  174, 200  417.4
Since 90° – 32° = 58°, the original bearing
could have been S58ºE. e2  d 2  f 2  2df cos E
e2  4002  802  2(400)(80) cos83
46 First, make a diagram.
 158,600
e  158.600  398.2
The guy wire anchored downhill is about 417.4 feet
long. The one anchored uphill is about 398.2 feet
long.
48.
a. Using the figure,

B  90  45  45
b2  a 2  c 2  2ac cos B
b2  122  302  2(12)(30) cos 45  535
b  535  23.1 In the figure, b = the guy wire anchored downhill, e =
You are about 23.1 miles from the pier. the guy wire anchored uphill.
B  90  5  95
a b
b.  E  90  5  85
sin A sin B
b2  a 2  c 2  2ac cos B
12 535
 b2  2002  1502  2(200)(150) cos 95  67,729
sin A sin 45
535 sin A  12sin 45 b  67,729  260.2
12sin 45 e2  d 2  f 2  2df cos E
sin A   0.3669
535 e2  2002  1502  2(200)(150) cos85  57, 271
A  22
Since 90  22  68 , the original bearing e  57, 271  239.3
could have been S68°E. The guy wire anchored downhill is about 260.2 feet
long. The one anchored uphill is about 239.3 feet
47. long.

49. 52. First, find the area using Heron’s formula.

1 1
s  ( a  b  c)  (320  510  410)  620
2 2
Area  s ( s  a )( s  b)( s  c )
 620(620  320)(620  510)(620  410)
 4, 296,600,000  65,548.46
Now multiply by the price per square foot.
(65, 548.46)(4.50)  294, 968
The cost is $294,968, to the nearest dollar.
Using the figure, 53. – 60. Answers may vary.

B  90  2  45 (using symmetry)
61. does not make sense; Explanations will vary.
b2  a 2  c 2  2ac cos B
Sample explanation: The Law of Cosines is not
b2  902  60.52  2(90)(60.5) cos 45 simply the negative of the Law of Sines.
 4060
62. makes sense
b  4060  63.7
It is about 63.7 feet from the pitcher’s mound to first 63. makes sense
base.
64. makes sense
50.
65.
Using the given information and the hint, we arrive at

the figure above. Let a = the side opposite the 35°
Using the figure, angle, c = the side opposite the 145° angle.
B  90  2  45 (using symmetry)
a 2  152  102  2(15)(10) cos 35  79.3
b2  a 2  c 2  2ac cos B
a  79.3  8.9
b2  602  462  2(60)(46) cos 45  1813
c 2  152  102  2(15)(10) cos145  570.7
b  1813  42.6
b  570.7  23.9
It is about 42.6 feet from the pitcher’s mound to third
The lengths of the parallelogram’s sides are about 8.9
base.
inches and 23.9 inches.
51. First, find the area using Heron’s formula.
66. If we call the lower left point D, and the lower right
1 1
s  ( a  b  c)  (240  300  420)  480 point E, then the Law of Cosines will give all three
2 2 angles in triangle ADE and triangle ABE. That
Area  s ( s  a )( s  b)( s  c ) allows us find A  29, B  87, and C  64. The
 480(480  240)(480  300)(480  420) Law of Sines will then allow us to find
a  11.6 and b  23.9.
 1, 244,160,000  35, 272.65
Now multiply by the price per square foot.
(35, 272.65)(3.50)  123, 454
The cost is $123,454, to the nearest dollar.

67. Section 7.3
Check Point Exercises
1. a. ( r , θ )  (3, 315)
Because 315° is a positive angle, draw
The angle between the minute and hour hand is
2
of θ  315 counterclockwise from the polar axis.
3 Because r > 0, plot the point by going out 3
the 90° angle from 9 to 12, or 60°. units on the terminal side of θ .
Let d = the distance between the tips of the hands.
d 2  m 2  h 2  2mh cos 60
1
 m 2  h 2  2mh  
2
 m 2  h 2  mh
d  m 2  h 2  mh
68. Answers may vary.

b. ( r , θ )  ( 2, π )
69. y  3 is a horizontal line through (0, 3).
Because π is a positive angle, draw θ  π
counterclockwise from the polar axis. Because r
< 0, plot the point by going out 2 units along the
ray opposite the terminal side of θ .
70. x 2  ( y  1)2  1 is a circle centered at (0, 1) with a

radius of 1.
 π
c. ( r , θ )   1,  
 2
71. x2  6 x  y 2  0 π π
Because  is a negative angle, draw θ  
2 2
x  6x
2
y 0
2
clockwise from the polar axis. Because r < 0,
x2  6x  9  y2  0  9 plot the point by going out one unit along the
( x  3)2  y 2  9 ray opposite the terminal side of θ .
( x  3)2  y 2  9 is a circle centered at (3, 0) with

a radius of 3.

Section 7.3 Polar Coordinates
2. a. Add 2π to the angle and do not 4.

change r.
 π  π   π 8π 
 5,    5,  2π    5,  
4 4 4 4 
 9π 
  5, 
 4 
b. Add π to the angle and replace r by –r.

 π  π   π 4π 
 
2
 5,    5,  π    5,   r x 2  y 2  12   3
4 4 4 4 
 5π   1 3  4  2
  5, 
 4  y  3
tan θ    3
x 1
c. Subtract 2π from the angle and do not π
change r. Because tan  3 and θ lies in quadrant IV,
3
 π  π   π 8π  π 6π π 5π
 5,    5,  2π    5,   θ  2π    
4 4 4 4  3 3 3 3

  5, 

7π 
 
The polar coordinates of 1,  3 are 
4 
 5π 
( r , θ )   2, 
3. a. ( r , θ )  (3, π )  3 
x  r cos θ  3cos π  3(1)  3
5.
y  r sin θ  3sin π  3(0)  0
The rectangular coordinates of (3, π ) are
(–3, 0).
 π
b. ( r , θ )   10, 
 6
π  3
x  r cos θ  10 cos  10  
6  2 
 5 3
π r  x 2  y 2  (0)2  ( 4)2  16  4
1
y  r sin θ  10sin  10    5 The point (0, –4) is on the negative y-axis. Thus,
6 2
3π  3π 
 π θ . Polar coordinates of (0, –4) are  4,  .
The rectangular coordinates of  10,  are 2  2 
 6
5 
3, 5 . 6. a. 3x  y  6
3r cos θ  r sin θ  6
r (3cos θ  sin θ )  6
6
r
3cos θ  sin θ

b. x 2  ( y  1)2  1 Concept and Vocabulary Check 7.3
 r cos θ 2  (r sin θ  1)2  1 1. pole; polar axis

r cos θ  r sin θ  2r sin θ  1  1
2 2 2 2
2. pole; polar axis
r 2  2 r sin θ  0
3. II
r  r  2sin θ   0
r  0 or r  2sin θ  0 4. IV
r  2sin θ 5. IV
7. a. Use r 2  x 2  y 2 to convert to a rectangular 6. III

equation.
r4 7. IV
r 2  16 8. II
x  y  16
2 2
9. r
The rectangular equation for r = 4 is
x 2  y 2  16. 10. r
11. r cos θ ; r sin θ

y
b. Use tan θ  to convert to a rectangular
x 12. squaring; x 2  y 2
equation in x and y.
3π
θ y
4 13. tangent;
x
3π
tan θ  tan
4 14. multiplying ; r; x 2  y 2 ; y
tan θ  1
y
 1
x Exercise Set 7.3
y  x
1. 225º is in the third quadrant.
3π C
The rectangular equation for θ  is
4
y   x. 2. 315° is in the fourth quadrant.
D
c. r  2sec θ
2 5π
r 3.  225 is in the third quadrant. Since r is
cos θ 4
negative, the point lies along the ray opposite the
r cos θ  2
terminal side of θ , in the first quadrant.
x  2 A
d. r  10sin θ π
4.  45 is in the first quadrant. Since r is negative,
r  10r sin θ
2
4
x 2  y 2  10 y the point lies along the ray opposite the terminal side
of θ , in the third quadrant.
x 2  y 2  10 y  0 C
x 2  y 2  10 y  25  25
5. π  180 lies on the negative x-axis.
x 2  ( y  5)2  25 B

6. 0 = 0° lies on the positive x-axis. Since r is negative, 13. Draw θ  90 counterclockwise, since θ is positive,
the point lies along the ray opposite the terminal side from the polar axis. Go out 3 units on the terminal
of θ , on the negative x-axis. side of θ , since r > 0.
B
7. –135° is measured clockwise 135° from the positive

x-axis. The point lies in the third quadrant.
C
8. –315° is measured clockwise 315° from the positive

x-axis. The point lies in the first quadrant.
A
3π
9.   135 is measured clockwise 135° from the
4 14. Draw θ  270 counterclockwise, since θ is
positive x-axis. Since r is negative, the point lies positive, from the polar axis. Go out 2 units on the
along the ray opposite the terminal side of θ , in the terminal side of θ , since r > 0.
first quadrant.
A
5π
10.   225 is measured clockwise 225° from the
4
positive x-axis. Since r is negative, the point lies
along the ray opposite the terminal side of θ , in the
fourth quadrant.
D
11. Draw θ  45 counterclockwise, since θ is positive, 4π

from the polar axis. Go out 2 units on the terminal 15. Draw θ   240 counterclockwise, since θ is
side of θ , since r > 0. 3
positive, from polar axis. Go out 3 units on the
terminal side of θ , since r > 0.
12. Draw θ  45 counterclockwise, since θ is positive,

from the polar axis. Go out 1 unit on the terminal side
of θ , since r > 0.

7π π
16. Draw θ   210 counterclockwise, since θ is 19. Draw θ    90 clockwise, since θ is positive,
6 2
positive, from the polar axis. Go out 3 units on the from the polar axis. Go 2 units out on the ray
terminal side of θ , since r > 0. opposite the terminal side of θ , since r < 0.
17. Draw θ  π  180 counterclockwise, since θ is 20. Draw θ  π  180 clockwise, since θ is
positive, from the polar axis. Go one unit out on the negative, from the polar axis. Go 3 units out on the
ray opposite the terminal side of θ , since r < 0. ray opposite the terminal side of θ , since r < 0.
3π π
18. Draw θ   270 counterclockwise, since θ is 21. Draw θ   30 counterclockwise, since θ is
2 6
positive, from the polar axis. Go one unit out on the positive, from the polar axis. Go 5 units out on the
ray opposite the terminal side of θ , since r < 0. terminal side of θ , since r > 0.
a. Add 2π to the angle and do not

change r.
 π  π   13π 
 5,    5,  2π    5, 
6 6 6 

b. Add π to the angle and replace r by –r. a. Add 2π to the angle and do not
 π  π   7π  change r.
 5,    5,  π    5,   3π   3π   11π 
6 6 6 
10,   10,  2π   10,
  
4 4 4 
c. Subtract 2π from the angle and do not change
r. b. Add π to the angle and replace r by –r.
 π  π   11π   3π   3π   7π 
 5,    5,  2π    5,   10,    10,  π    10, 
6 6 6  4 4   4 
π
22. Draw θ   30 counterclockwise, since θ is r.
6  3π   3π   5π 
positive, from the polar axis. Go out 8 units on the 10,   10,  2π   10,
  
4 4 4 
2π
24. Draw θ   120 counterclockwise, since θ is
3
positive, from the polar axis. Go out 12 units on the
a. Add 2π to the angle and do not change r.

 π  π   13π 
 8,    8,  2π    8, 
6 6 6 

 π  π   7π  a. Add 2π to the angle and do not change r.
 8,    8,  π    8,   2π   2π   8π 
6 6 6 
12,   12,  2π   12, 
3   3   3 
r. b. Add π to the angle and replace r by –r.
 π  π   11π   2π   2π   5π 
 8,    8,  2π    8,   12,    12,  π    12,
  
6 6 6  3 3 3 
3π c. Subtract 2π from the angle and do not change r.

23. Draw θ   135 counterclockwise, since θ is 2π   2π 4π 
4   
12,   12,  2π   12, 
  
positive, from the polar axis. Go out 10 units on the 3 3 3 

π c. Subtract 2π from the angle and do not change

25. Draw θ   90 counterclockwise, since θ is r.
2
positive, from the polar axis. Go 4 units out on the  π  π   3π 
terminal side of θ , since r > 0.  6,    6,  2π    6,  
2 2 2 
27. a, b, d
28. a, c, d
29. b, d
30. a, d
31. a, b
32. a, c
a. Add 2π to the angle and do not 33. The rectangular coordinates of (4, 90°)
change r. are (0, 4).
 π  π   5π 
 4,    4,  2π    4,  x  r cos θ  6 cos180  6(1)  6
2 2 2  34.
y  r sin θ  6sin180  6  0  0
b. Add π to the angle and replace r by –r. The rectangular coordinates of (6, 180°) are
 π  π   3π  (–6, 0)
 4,    4,  π    4, 
2 2 2 
π 1
35. x  r cos θ  2 cos  2   1
c. Subtract 2π from the angle and do not change 3 2
r.  3
π
 π  π   3π  y  r sin θ  2 sin  2  3
 4,    4,  2π    4,   3  2 
2 2 2 
 π
The rectangular coordinates of  2, 
26. Draw θ  π  90 counterclockwise, since θ is  3
positive, from the polar axis. Go 6 units out on the
terminal side of θ , since r > 0. 
are 1, 3 . 
π  3
36. x  r cos θ  2 cos  2  3
6  2 
π 1
y  r sin θ  2 sin  2   1
6 2
 π
The rectangular coordinates of  2,  are
 6
 
3,1 .
π
37. x  r cos θ  4 cos  4  0  0
2
a. Add 2π to the angle and do not change r. π
y  r sin θ  4 sin  4(1)  4
 π  π   5π  2
 6,    6,  2π    6, 
2 2 2   π
The rectangular coordinates of  4, 
 2
are (0, –4).
 π  π   3π 
 6,    6,  π    6, 
2 2 2

3π
 
2
38. x  r cos θ  6 cos  6  0  0 43. r x 2  y 2  (2)2  2 3
2
3π  4  12  16  4
y  r sin θ  6sin  6(1)  6
2 y 2 3
 3π  tan θ    3
The rectangular coordinates of  6,  are x 2
 2  π
(0, 6). Because tan  3 and θ lies in quadrant IV,
3
x  r cos θ  7.4 cos 2.5  7.4(0.80)  5.9 π 5π
39. θ  2π   .
3 3
y  r sin θ  7.4 sin 2.5  7.4(0.60)  4.4
The rectangular coordinates of (7.4, 2.5) are 
The polar coordinates of 2, 2 3 are 
approximately (–5.9, 4.4).  5π 
( r , θ )   4, .
40. x  r cos θ  8.3cos 4.6  8.3( 0.11)  0.9  3 
y  r sin θ  8.3sin 4.6  8.3( 0.99)  8.2
 2 3 
2
The rectangular coordinates of (8.3, 4.6) are 44. r x2  y2   (2)2
approximately (–0.9, –8.2).
 12  4  16  4
41. r x  y  ( 2)  2
2 2 2 2
tan θ 
y

2

1
x 2 3 3
 44  8  2 2
π 1
y 2 Because tan  and θ lies in quadrant II,
tan θ    1 6 3
x 2
Because tan θ  1 and θ lies in quadrant II, π 5π
θ π  .
3π 6 6
θ
4
.

The polar coordinates of 2 3, 2 are 
The polar coordinates of  2, 2 are
 5π 
( r , θ )   4, .
 3π   6 
( r , θ )   8, .
 4 
 3 
2
45. r x2  y2   (1)2
42. r x  y  (2)  (2)
2 2 2 2
 31  4  2
 44  8  2 2
y 1 1
y 2 tan θ   
tan θ    1 x  3 3
x 2
π π 1
Because tan  1 and θ lies in quadrant IV, Because tan  and θ lies in quadrant III,
4 6 3
π 7π π 7π
θ  2π   . θ π  .
4 4 6 6
The polar coordinates of (2, –2) are 
The polar coordinates of  3,  1 are 
 7π   7π 
( r , θ )   8,  or  2 2 , .  7π 
 4   4  ( r , θ )   2, .
 6 

x7
 
2 51.
46. r x 2  y 2  ( 1) 2   3
r cos θ  7
 1 3  4  2 7
r
y  3 cos θ
tan θ    3
x 1
52. y3
π
Because tan 3 and θ lies in quadrant III, r sin θ  3
3
π 4π 3
θ π  r
3 3
. sin θ

The polar coordinates of 1,  3 are  53. x2  y 2  9
 4π  r2  9
( r , θ )   2,
 3 
.
r3
47. r x 2  y 2  (5)2  (0)2  25  5 54. x 2  y 2  16

y 0 r 2  16
tan θ   0
x 5 r4
Because tan 0  0 and θ lies on the polar axis,
θ 0. 55. ( x  2) 2  y 2  4
The polar coordinates of (5, 0) are
( r , θ ) = (5, 0). ( r cos θ  2)2  ( r sin θ )2  4
r 2 cos2 θ  4r cos θ  4  r 2 sin 2 θ 2  4
48. r x  y  (0)  ( 6)  36  6
2 2 2 2
r 2  4r cos θ  0
y 6 r 2  4r cos θ
tan θ    undefined
x 0 r  4 cos θ
π
Because tan is undefined and θ lies on the
2 56. x 2  ( y  3)2  9
π 3π
negative y axis, θ  π  . ( r cos θ )2  ( r sin θ  3)2  9
2 2
r 2 cos2 θ  r 2 sin 2 θ  6r sin θ  9  9
 3π 
The polar coordinates of (0, –6) are ( r , θ )   6,
 2 
.
r 2  6r sin θ  0
r 2  6r sin θ
49. 3x  y  7
r  6sin θ
3r cos θ  r sin θ  7
r (3cos θ  sin θ )  7 57. y2  6x
7 ( r sin θ )2  6r cos θ
r
3cos θ  sin θ
r 2 sin 2 θ  6r cos θ
50. x  5y  8 r sin 2 θ  6cos θ
r cos θ  5r sin θ  8 6cos θ
r
r (cos θ  5sin θ )  8 sin 2 θ
8
r
cos θ  5sin θ

58. x2  6 y π
62. θ
( r cos θ )2  6r sin θ 3
π
r 2 cos 2 θ  6r sin θ tan θ  tan
3
r cos 2 θ  6sin θ tan θ  3
6sin θ y
r  3
cos 2 θ x
59. r8 y  3x
r  64
2
x  y 2  64
2
63. r sin θ  3
y3
60. r  10
r  100
2
x  y 2  100
2
64. r cos θ  7
x7
π
61. θ
2
π
tan θ  tan
2
tan θ is undefined 65. r  4 csc θ
y 4
is undefined r
x sin θ
x =0 r sin θ  4
y4

66. r  6sec θ 69. r  12 cos θ
r
6 r  12 r cos θ
2
cos θ
x  y 2  12 x
2
r cos θ  6
x 2  12 x  y 2  0
x6
x 2  12 x  36  y 2  36
 x  62  y 2  36
67. r  sin θ
r  r  r  sin θ
r 2  r sin θ
70. r  4 sin θ
x2  y2  y
r  4r sin θ
2
x  y 2  4 y
2
x2  y 2  4 y  0
x2  y 2  4 y  4  4
x2   y  2  4
2
68. r  cos θ
r  r  r  cos θ
r 2  r cos θ
x2  y 2  x
71. r  6 cos θ  4 sin θ

r  r  r (6 cos θ  4 sin θ )
r 2  6r cos θ  4r sin θ
x2  y 2  6 x  4 y

72. r  8cos θ  2 sin θ 76. r  a csc θ

r  r  r (8cos θ  2 sin θ ) a
r
r  8r cos θ  2r sin θ
2 sin θ
r sin θ  a
x  y 2  8x  2 y
2
ya
This is the equation of a horizontal line.
77. r  a sin θ
r  ar sin θ
2
x  y 2  ay
2
x 2  y 2  ay  0
a2 a2
73. r 2 sin 2θ  2 x 2  y 2  ay  
4 4
r 2 (2 sin θ cos θ )  2 2 2
 a a
2r sin θ r cos θ  2 x2   y     
 2 2
2 yx  2
a
xy  1 This is the equation of a circle of radius centered
2
1  a
y at  0,  .
x  2
78. r  a cos θ
r 2  ar cos θ
x 2  y 2  ax
x 2  ax  y 2  0
a2 a2
x 2  ax   y2 
4 4
74. r 2 cos 2θ  2 2 2
 a a
 x  2   y   2 
2
r (cos θ  sin θ )  2
2 2 2
r 2 cos2 θ  r 2 sin 2 θ  2 a
This is the equation of a circle of radius centered
( r cos θ )  ( r sin θ )  2
2 2 2
x2  y2  2 a 
at  ,0  .
2 
 π
79. r sin θ    2
 4
 π π
r  sin θ cos  cos θ sin   2
 4 4
2 2
r sin θ   r cos θ  2
2 2
75. r  a sec θ
2 2
a y  x 2
r 2 2
cos θ
y  x2 2
r cos θ  a
xa y  x  2 2 has slope of 1 and y-intercept of 2 2 .
This is the equation of a vertical line.

 π 83. The angle is measured counterclockwise from the

80. r cos θ    8 polar axis.
 6
2 4π
 π π θ  (360)  240 or .
r  cos θ cos  sin θ sin   8 3 3
 6 6 The distance from the inner circle’s center
π π to the outer circle is
r cos θ cos  r sin θ sin
8
6 6 r  6  3(3)  6  9  15
3 1  4π 
x  y 8 The polar coordinates are ( r , θ ) = 15, .
2 2  3 
x 3  y  16
84. The angle is measured counterclockwise from the
 y   x 3  16 5 5π
polar axis. θ  (360)  300 or .
y  x 3  16 6 3
On the inner circle, r = 6.
y  x 3  16 has slope of 3 and y-intercept of –16.
 5π 
The polar coordinates are ( r , θ )   6,
 3 
.
2π
81. x1  r cos θ  2 cos  1
3 85. (6.3, 50°) represents a sailing speed of 6.3 knots at an
2π angle of 50° to the wind.
y1  r sin θ  2 sin  3
3
1, 3 
86. (7.4, 85°) represents a sailing speed of 7.4 knots at an
angle of 85° to the wind.
π 87. Out of the four points in this 10-knot-wind situation,
x2  r cos θ  4 cos 2 3
6 you would recommend a sailing angle of 105°. A
π sailing speed of 7.5 knots is achieved at this angle.
y2  r sin θ  4 sin 2
6 88. – 96. Answers may vary.
 2 3, 2  97.
d  x2  x1  2
  y2  y1 
2
2   
2 2
d 3 1  2  3
d2 5
To three decimal places, the rectangular coordinates
82. x1  r cos θ  6 cos π  6 are (–2, 3.464).
y1  r sin θ  6sin π  0 98.
 6, 0
7π 5 2
x2  r cos θ  5cos 
4 2
7π 5 2
y2  r sin θ  5sin  To three decimal places, the rectangular coordinates
4 2 are (–0.670, 5.157).
5 2 5 2 
 2 , 2  99.
 
d  x2  x1 2   y2  y1 2
2 2
5 2   5 2 
d   6     0
 2   2  To three decimal places, the rectangular coordinates
are (–1.857, –3.543).
d  61  30 2

100.
To three decimal places, the polar coordinates are ( r , θ ) = (5.385, 2.761).
101.
To three decimal places, the polar coordinates are ( r , θ ) = (3, 0.730).
102.
To three decimal places, the polar coordinates are ( r , θ ) = (8.674, –2.091).

Adding 2π to –2.090514401 will give the approximate equivalent angle 4.193, or polar coordinates (8.674, –2.091).
103. does not make sense; Explanations will vary. Sample explanation: There are multiple polar representations for a given
point.
104. does not make sense; Explanations will vary. Sample explanation: There is only one rectangular representation for a
given point.
105. makes sense
106. makes sense
107. Use the distance formula for rectangular coordinates, d  ( x2  x1 )2  ( y2  y1 )2 .

Let x1  r1 cos θ1 , y1  r1 sin θ1 ,
x2  r2 cos θ 2 , y2  r2 sin θ 2
d  r2 cos θ 2  r1 cos θ1 2   r2 sin θ 2  r1 sin θ1 2

 r22 cos 2 θ 2  2 r1 r2 cos θ1 cos θ 2  r12 cos2 θ1  r22 sin 2 θ 2  2 r1 r2 sin θ1 sin θ 2  r12 sin 2 θ1
   
 r22 cos 2 θ 2  sin 2 θ 2  r12 cos2 θ1  sin 2 θ1  2r1 r2 cos θ1 cos θ 2  sin θ1 sin θ 2 
 r22 (1)  r12 (1)  2r1 r2  cos θ 2  θ1 
 r12  r22  2r1 r2 cos θ 2  θ1 

 5π   π
108. Let ( r1 , θ1 )   2,  , ( r2 , θ 2 )   4, 
 6  6
d  r12  r22  2r1 r2 cos(θ 2  θ1 )
 π 5π 
 22  42  2(2)(4) cos  
 6 6 
 2π 
 20  16cos 
 3 
 1
 20  16     20  8  28 or 2 7
 2
π π π 2π 5π
θ 0 π
6 3 2 3 6
109. 2 3 1 3 2 3
r  1  cos θ 0 2 2 1 2 2 0
 0.13  0.5  1.5  1.87
π π π 2π 5π 7π 4π 3π
θ 0 π
6 3 2 3 6 6 3 2
110.
1 3 1 3 1 3
r  1  2 sin θ 1 2 3 2 1 0 1
 2.73  2.73  0.73

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commercial intercourse were about to commence, as it was
supposed the Chinese, as of old, would soon begin to form pepper-
plantations, and the expectation was partly fulfilled. A rich
shopkeeper obtained permission from the sultan, and a grant of land
having been made, he set to work to form a garden. He planted
fruits, vegetables, and pepper, the last grew luxuriantly, though the
soil appeared unpromising; but no sooner was it known to be
yielding, than crowds of idlers from the capital flocked there, and
soon stripped it of everything eatable. In despair, he gave up his
project, and no one has had the courage to try again; in fact, it would
be useless as long as the present system of government holds.
Since our colony of Labuan was established in 1848, a few
Chinese have left it to spread along the coast. As yet they have had
little effect, since most of them have married into native families, and
done little else than carry on a petty trade, or manufacture arrack.
Lately, however, a few have commenced pepper-gardens, but the
districts are too unsettled to promise much, yet it is a movement to
be encouraged. When I have asked the Chinese why they did not
emigrate there, the invariable answer has been, “Will you afford us
protection? if not, as soon as our plantations are productive, the
nobles will force the produce from us.”
The sultan, in a moment of enlightenment, determined to
encourage planters, and actually sent for a dozen from Singapore,
paid their expenses to Brunei, and promised a monthly allowance till
the produce of their vegetable-gardens enabled the Chinese to
support themselves. These regular payments, however, soon
became very distasteful to him, and every month be gave them less,
till at last from want of food they all dispersed, and the experiment
has not been tried again.
The Brunei government makes every effort to prevent the Chinese
mixing with the aborigines, as it tends to destroy the monopoly of
trade they seek to establish, and they fear also the teaching of the
Chinese, who would never counsel submission to oppressive rulers,
though when employed by the nobles as agents, they can be more
systematically grinding than the Malays.
It is evident that the intercourse between Borneo and China,
which undoubtedly was once very active, has been decreasing for
above a hundred years, and the cause was doubtless the anarchy
into which the country fell and the consequent want of protection.
Fifty years ago, the junk trade appears entirely to have ceased; and
even in 1775 it had been reduced to about seven a year, although
they continued to build vessels at Brunei.
With regard to the accounts of old travellers, that the north of
Borneo was formerly peopled from Cochin China, I have heard
nothing to support the theory, beyond the tradition that in ancient
days a great trade was carried on between Annam and the north-
west coast, when many Cochin Chinese settled in Borneo. In fact, in
the Champa country, in the southern portion of the Kambodian
peninsula, there is a people whose language contains a
considerable number of Malay words; so that the effect on these two
countries may have been mutual, though Champa, doubtless, was
more influenced by settlers from the Malay peninsula.
I have before alluded to the Chinese wandering from our colony of
Labuan to settle in small numbers in the districts on the coast to the
north of that island. A favourite place was Papar, as the Dusuns
there were wealthy, and, being numerous, cared little for their
nominal ruler, pañgeran Omar; and among the Chinese settlers were
two men, who lived in a small house on the banks of the river. One
day, early in the year 1859, the wife of the chief of a neighbouring
village was passing that way, when one of these men attempted to
pull off her petticoat, which constituted her only covering. Her
screams bringing some friends to the spot, the man let her go and
ran into his house. In the evening, the chief came to demand
satisfaction for this very gross insult, but said, as the offender was a
stranger, and perhaps did not know the customs of the country, he
should only fine him the value of a goat. The two Chinese ordered
him to leave their house, and, to enforce it, took up their carrying-
sticks, with which the one who had insulted the woman struck him.
The Dusun chief, who had his spear in his hand, stabbed the
offender and wounded his companion, who came up to join in the
attack. This affair caused great commotion in the district, and all the
Chinese clamoured for vengeance.
Pañgeran Omar inquired into the case, fined the Dusun, and
ordered the amount to be paid over to the dead man’s friends. They
were not, however, satisfied with the amount of the fine, and
determined to revenge themselves. Collecting a body of about
twenty of their countrymen, on the pretence of a pig hunt, they
marched to attack the chief’s village; upon which the Dusuns,
beating the alarm gong, soon apprised their neighbours that they
were in danger; and the Chinese, as usual, arrogant when there was
no opposition, but cowards in circumstances of peril, immediately on
finding their first volley did not frighten their enemies, fled with
precipitation, and were pursued by the Dusuns, and the larger
portion of them killed.
The case was misrepresented in Labuan, and some demands
were made for satisfaction; but it was evident the Chinese had
brought this disaster on themselves; and I know of no worse policy
than to consider all those, whether British subjects or not, who leave
our colony to settle on the coast as entitled to our protection. If we
can be of service to them, it is as well to use our influence to insure
them the best treatment, but we should never let the Chinese
imagine we intend to give them the protection of the British flag on all
occasions. Yet it is a subject which requires delicate handling, for, if
we entirely abandoned their interests, they would be plundered and
massacred; and without them there will never be any progress on
the coast, or developement of trade and agriculture on a large scale;
and if we claim them as British subjects, which a few are in reality,
their insolence to the natives is often unbearable.
I have generally found that those Chinese who come direct from
their own country are better adapted to succeed with the native
chiefs than those who have resided long in our own settlements,
where they acquired an independence almost amounting to
lawlessness. I once nearly lost my life through the reckless conduct
of one of these Singapore Chinese, who had been accustomed to
treat the Malays there with great contumely. When he arrived in
Brunei, he did the same thing with a crazy man belonging to the
Pablat section of the town, and the Chinese quarter was thrown into
confusion. I sent both men to the sultan, but in the meantime the
report spread among the Malay’s relatives that the Chinese had ill
used him, and 150 men immediately came down, shouting that they
would run amuck among the Chinese. A respectable Bornean trader
came hastily into my room, saying, if I did not immediately go down
to the scene, there would be a massacre. I caught up my sword and
hurried to the Chinese village, to find the Pablat men in the act of
assaulting the strangers; and had one wound been given, there
would have been no stopping the mischief.
I need not dwell on all the particulars, but it was with the greatest
difficulty I turned the Malays back from their purpose. To me they
behaved with great civility, after the first excitement was over; but the
glare their chief gave me, when I put the hilt of my sword to his
breast to prevent him using his spear on an unfortunate Chinese
trader, who had nothing to do with the quarrel, was a very savage
one. His hand in a moment sought his kris; but on my saying, in a
very quiet tone, “Don’t draw your kris on me,” he dropped his
intention at once, and although his followers drew their weapons and
urged him to the attack, he began to explain to me the reason of his
coming with that force at his back. I knew if I could check the rush for
five minutes, things would be safe, as by that time some friends, who
were staying at the house, would be down with all my armed
followers; and so it proved. But the insolence of the Chinese was
effectually checked by this demonstration, and I had no further
trouble with them, as they thought I might not always be there to
stand between them and death.
This is but a meagre account of the results of that extensive
Chinese intercourse with the northern portion of Borneo, which was
carried on for so many hundred years; but in a country so uncivilized
there are no antiquities; and although the tradition exists among the
people that formerly numerous immigrants arrived and settled, still
they can relate few facts concerning them. There can be little doubt,
judging from the character of the two people, that the nobles would
endeavour to squeeze out of the foreign planters as much as
possible; that they would fine them heavily for very slight faults, till
they would drive the Chinese to resistance, and insurrections would
as surely follow among a people who always unite against other
races. They are no match for the Malays and Dayaks in wild warfare;
and it is only their organization which enables them to offer any
resistance to the desultory attacks of their enemies.
It has been said that in the great insurrection of the Chinese the
Muruts joined them, and that the Borneans were compelled to seek
the assistance of the Sulus to repress it, but I did not hear any
mention of the latter statement, and it appears improbable. Internal
dissention is the more likely cause of the failure of the attempt to
throw off the yoke of the Malays, the Muruts being bribed to leave
their allies. At all events, the Bisaya tribes were engaged in its
suppression, as the grandfather of the orang kaya of Blimbing
assisted in taking the fort at the entrance of the Madalam river. The
Chinese insurgents, driven from the lower country, attempted to
make a stand on a rounded hill there, but lost their fort, either by a
panic or by treachery, my informant did not appear certain which.
Between Brunei and Sarawak the Chinese do not appear to have
established themselves; but to the latter country the gold-workers of
Sambas occasionally sent parties of men to try the soil, as auriferous
ore was reported to be plentiful. But during the distractions
consequent on the civil war, they found it impossible to pursue their
peaceful industry, and those who were successful in obtaining gold
were exposed to the attacks of lawless Malays.
One man, who is now a very respected member of society, a haji
of mark, who has for the last twenty years conducted himself in the
most exemplary manner, was once tempted to commit a crime by the
report that a party of Chinese was returning to Sambas with sixty
ounces of gold. He and a few of his relations waylaid the travellers,
and, surprising them in the dark forest, murdered them and obtained
the treasure.
This naturally aroused the anger of their countrymen, and an
expedition was fitted out at Sambas to revenge the deed. They
marched into the Sarawak territory, and advanced nearly as far as
the town of Siniawan, then occupied by Malays, but found a strong
stockade built across the path. The Chinese numbered about seven
hundred men, while their opponents were at first scarcely twenty, but
protected by their position and numerous guns. Confident in their
numbers, the assailants rushed to the attack, almost reaching the
foot of the defences, but receiving a severe fire from the guns in
position, loaded with nails, bits of old iron and shot, they were beaten
back. The Malays acknowledge the Chinese kept up the attacks all
day; but, after their first repulse, they principally confined themselves
to a distant fire, though they occasionally made attempts to turn the
position, but were repulsed by the ever-increasing numbers of the
Malays.
Towards evening the Chinese withdrew to the banks of the river,
and made preparations to pass the night; while the Malays, who had
been reinforced by many of their friends, determined to try the effect
of a surprise. They were commanded by the gallant patinggi Ali,
whose exploits and death are recorded in the Voyage of the Dido;
and just at sunset they started in their light boats with a gun in each,
and pulled with an almost silent stroke towards the Chinese
encampment, where they found their enemies cooking rice, smoking
opium, or shouting or talking, in fact, making so great a noise as to
prevent the possibility of hearing the sound of paddles, cautiously
pulled.
When all were ready, patinggi Ali gave the signal to fire, and the
next moment they yelled and sprang ashore. The startled Chinese
fled, and were pursued relentlessly by the Dayaks, who had come
down from the hills to share in the struggle. It is said half the
invading force was destroyed, and that the old, dried skulls I had
noticed in the Dayak villages were the trophies of the fight.
Being thus exposed to every kind of ill-treatment, it is not
surprising that the Chinese did not care to settle in the country; but,
after Sir James Brooke was established in Sarawak, they began to
increase in numbers, though always inclined to be troublesome.
When I arrived, in the year 1848, it was considered there were about
six hundred living there, mostly engaged in gold-working, and even
these were much inclined to have an imperium in imperio, though too
weak to carry out their views. They had formed themselves into an
association called the Santei Kiu kunsi, or company.
I must notice that these Chinese are not the pure emigrants from
China, but the half-breeds, descendants of the early settlers, who
obtained Malay and Dayak wives, and are more warlike in their
habits than the pure Chinese, and many have much of the activity of
the aborigines. Settled in Sambas before the arrival of the
Europeans in those seas, they gradually formed self-governing
communities among the weak Malay States around, and by
intermarriage with the women of the Dayak tribes in their
neighbourhood, formed both political and social alliances with them.
It was not to be expected that this state of things could long exist
without serious disputes arising with the Malay chiefs; however, they
generally managed to prevent a total estrangement; but when
backed by the Dutch officials, the Sultan of Sambas endeavoured to
coerce them into submission, the Chinese gold-working communities
refused to obey, attacked the small force sent against them,
captured the forts, and drove the Dutch troops to their steamers, and
left in their possession little more than the town of Sambas. This, of
course, roused the officials, and a strong expedition was sent from
Java, which within a year subdued the refractory Chinese, who, in
fact, submitted with very little opposition. But during their success
they managed to give Sarawak a considerable lift.
At the mouth of the Sambas river there is a place called
Pamañgkat, where several thousand Chinese agriculturists were
engaged in raising fine crops of rice. These men had not joined their
countrymen in their resistance to the Sambas Government, and were
therefore marked out for punishment during their brief success. In
their alarm, the Pamañgkat Chinese fled to Sarawak, arriving in
great numbers during the year 1850, whilst I was absent with Sir
James Brooke on his mission to Siam; some came by sea, others
fled overland to Lundu and to the interior.
I found on my return in October, after nearly a year’s absence,
that a great change had taken place in the appearance of the town of
Kuching: dozens of fresh houses were built and building, while the
surrounding forest was falling rapidly before the axes of the fugitives.
Many of them had arrived destitute of all property, and I learnt that
three hundred and ten families were entirely supported by the food
and money furnished by the Sarawak government, besides hundreds
of others having received presents of tools and temporary
assistance. It was calculated at the time that about three thousand
had arrived, many of whom immediately joined the gold-workers in
the interior. We found also that the mission school had received a
great addition in the form of about twenty remarkably intelligent-
looking little boys and girls, whose destitute parents had gladly
handed them over to the care of the clergy.
Sir James Brooke, hearing that there was much confusion in the
interior, from the numerous freshly-arrived Chinese, and from the
Dayaks being alarmed by this sudden influx into their
neighbourhood, started with a party to visit it. We soon reached
Siniawan, the little Chinese trading town I have previously described,
which was but now advancing to importance. We continued our
course up the river to Tundong, where there is a ghàt used by the
gold-working company to land their supplies. Nearly all the gold that
is worked in Borneo is done by kunsis, or companies, which
sometimes numbered several thousand men; in fact, they say that at
Montrado nearly the whole of its Chinese population and that of the
neighbourhood, estimated at 50,000 men, were included in one
kunsi. Generally, however, they consisted, as at Sarawak, of a few
hundred members, though they might still be in connexion with the
parent company. The great influx of Chinese had now, however,
swelled the Santei Kiu kunsi to inconvenient dimensions.
At Tundong we found a few store-houses and a very tolerable
path leading over to Bau, the principal Chinese settlement. The
views on either side of us, as we advanced, were sometimes
exceedingly picturesque; for, as we reached the summits of low hills,
a fine undulating country was spread out beneath us. The path led
through shady forests, then open Dayak clearings, along the sides of
hills, and over pretty streams spanned by very primitive bridges.
As we approached the town of Bau we met a band of Chinese
musicians who had come forth to greet us, and gun after gun was
discharged in honour of the rajah’s visit. Our procession was a very
motley one, half a dozen Englishmen, followed by a long line of
Malays, Chinese, and Dayaks, marching in Indian procession, some
carrying spears, others muskets, or flags.
At last we reached the kunsi’s house, prettily situated in the valley
of Bau, which was on two sides flanked by black-looking
perpendicular hills. The house itself was a substantial one, built of
ironwood posts and good planks, and roofed with excellent ironwood
shingles.
I will describe one of their gold-workings. They dammed up the
end of the valley at the back of the kunsi’s house, thus forming a
large reservoir of water, perhaps a quarter of a mile in length. The
dam was very neatly constructed, being completely faced with wood
towards the water, and partially on the outside, to enable it to resist
the very heavy rains which fall in this country. A ditch, about four feet
broad, was cut from the reservoir towards the ground which the
overlooker of the company had selected as a spot likely to produce a
good yield of gold, and a well-made sluice-gate was constructed in
the dam to supply the ditch with as much water as might be required;
minor sluice-gates to the main ditch enabled the smaller ones also to
receive supplies of water. When this was all prepared, the sluice-
gates were opened, and the earth in its neighbourhood thrown into
the ditch, and the rushing water carried off the mud and sand and
allowed the particles of gold to sink to the bottom. After three or four
months they cleaned out the ditch and carefully washed the residue,
which generally yielded them sufficient to make a tolerable division
among the workmen after all the expenses had been paid.
It is a very wasteful system of working gold; in fact, when we were
there, all the women and girls, lately arrived from Sambas, had the
privilege given them of washing the earth which had been swept
away by the rushing water, and I believe they obtained as much in
proportion to the number working as was divided among the men,
who had had all the labour of constructing these extensive works. No
one has yet taught them deep sinking; in fact, it is to be regretted
that none of their countrymen accustomed to the method of
procuring this precious metal in our Australian colonies have yet
visited Sarawak.
That there is an abundance of gold to be found there I verily
believe, and, as an instance, I may notice that in November, 1848, a
great landslip took place, and the face of the Trian mountain was laid
bare. Some Malays, observing small pieces of gold mixed with the
clay, began a strict search, and having great success, the news soon
spread, and several thousand people flocked to the spot, where they
worked till the heap of earth and stone was cleared away. All had fair
success, and we heard of none who got less than an ounce and a
half per month. The work lasted above six weeks. I saw one nugget
picked up, which weighed about seven ounces.
The influx of the Pamangkat Chinese gave great impetus to the
search for the auriferous ore, and new reservoirs, dams, and ditches,
were appearing in every direction; but yet the new-comers, being
only accustomed to agriculture, did not take very kindly to gold
digging. Sir James Brooke was anxious to remove a large body to
some district which they could cultivate; but they were too poor to be
able to support themselves while waiting for their crops. The gold
company was not willing to part with these people, and promised
them every assistance if they would stay at Bau.
Nearly all the early efforts to assist these immigrants in developing
the agricultural resources of the country had but little success. They
commenced rice farms at Si Jinkat on the Muaratabas, and also at
the foot of the Santubong hill; but though they were supplied with
food and tools by the Sarawak government, they abandoned both
attempts, and scattered themselves either among the gold-workers
in the interior, or removed to the district of Lundu, where, as I have
already mentioned, they made beautiful gardens. It was a matter of
regret that they should have abandoned Santubong, as the soil is of
a very fine description. I believe the non-success, however, arose
from defective management and inefficient superintendence.
Everything appeared to go on very quietly till January 1852, when
a fortnight’s continued rain rather injured their reservoirs, and laid the
country under water. Such a flood, they say, never before or since
came upon them. At Kuching it was necessary in the Chinese town
to move from house to house in boats. At Siniawan it rose to so
great a height that the inhabitants had to abandon their houses, and
an unfortunate Chinese, seeking safety in his garret, was drowned,
being unable to force his way through the roof; and up the country
we saw afterwards the dried grass left by the stream at least forty
feet above the usual level of the river.
In 1853, the gold company gave the government considerable
trouble, and had to be curbed by a great display of armed force; but
they submitted without any necessity of proceeding to extremities.
The case was this: the government had issued an order to the
company that they should not make any fresh reservoirs or gold-
workings among the Dayak lands without obtaining permission from
the authorities, as on several occasions quarrels had arisen between
them and the neighbouring tribes, on account of their taking
possession of the best farming ground in the country.
The Pamangkat Chinese were never quite satisfied with their
position as gold-workers, and constantly made applications to the
government for assistance in order to recommence their old style of
living as rice cultivators. At last they fixed on a good spot, and food
and rice were supplied to several hundreds. This well-managed
movement might have been increased to any extent, as all the late
immigrants preferred a quiet rural life; and by the commencement of
1856 nearly five hundred were established at a place called Sungei
Tañgah, about six miles above the town.
I may observe that during the four previous years the Dutch had
kept the Chinese within the boundary of their settlements in very
strict order; but, in 1856, some dispute taking place, a Dutch officer
and a party of troops were cut off by the workmen of the Lumar
kunsi, one of the large gold companies, about three hundred of
whom escaped over the borders into the Sarawak territories, while
the rest were captured, and many suffered condign punishment for
their crime.
I will notice here a regulation which obtains in the Dutch territories
of Sambas and the other border states, which is so illiberal that I can
scarcely believe it to be authorized by any of the superior authorities,
but must be the work of a very narrow-minded local official. No
Chinese, whether man, woman, or child, can leave the Dutch
territories without first paying a fine of 6l.; so that as very few
workmen can save that amount they are practically condemned to
remain there all their lives, unless they can evade the blockade kept
upon them, thus running the risk of the cat-o’nine-tails, a fine, and
imprisonment. The reason for this regulation is that no Chinese in
Borneo would willingly remain under Dutch rule who could possibly
escape from it; and if liberty were given to them to leave the country,
nearly every man would abandon it. Therefore, gunboats watch the
coast, and on the frontiers soldiers, Malays, and Dayaks, are
ordered to stop any Chinese who may attempt to escape from the
Dutch territories.
In the spring of 1856, I made a tour through the Chinese
settlements established in Sarawak, commencing with the rice
plantations and vegetable gardens established at Sungei Tañgah. I
have never seen in Borneo anything more pleasing to my eye than
the extensive cultivated fields which spread out around the scattered
Chinese houses, each closely surrounded by beds of esculent plants
growing in a most luxuriant manner.
Every day appeared to be adding to the area of cultivation;
because, as the agriculturists became more wealthy, they invited the
poorer gold-workers to join them, and were thus enabled to employ
many labourers. Already the effect of this increase of produce was
perceptible on prices, so that vegetables, fowls, and ducks, were
beginning to be bought at reasonable rates. On the other side, the
right-hand bank of the river, near the little mount of Stapok, about
forty Chinese had commenced gardens without any assistance from
Government, and appeared to be very prosperous. To this spot a
road had been cut through the forest from the town, which
afterwards became memorable in Sarawak annals. Altogether, as I
have before observed, there were about five hundred people
assembled here engaged in a war against the jungle.
Continuing our course towards the interior, we met with no
Chinese houses until we reached the village of Siniawan, at that time
governed by the guns of the little fort of Biledah, admirably situated
on a high point jutting into the river, and on the same spot where the
Sarawak Malays during the civil wars had their strongest stockade.
The town was remarkably flourishing, and we here heard a
confirmation of the reports that a great many Chinese were arriving
from Sambas. As we were anxious to be thoroughly acquainted with
the actual condition of the gold-working population, we determined to
walk across from Siniawan to the head-quarters of the gold company
at Bau. The paths were in very good condition, quite suited for riding
over, except when we reached the bridges thrown across the deep
gullies which intersect the country.
From Siniawan all the way to Bau, a distance of ten miles, there
was a constant succession of reservoirs and gold-workings; and
judging from the new houses springing up in every direction, we felt
sure the population was increasing. About a third of the way along
the road, a branch path led to a place called by the Chinese
“Shaksan,” where there was an excellent hot spring, over which Mr.
Ruppell had built a little house. We diverged to this spot to indulge in
the unusual sensation of a hot bath, and found the temperature of
the water so warm that it was almost unbearable; but for any one
suffering from rheumatism it would be excellent. We noticed in the
neighbourhood many limestone rocks water-worn into fantastic
shapes, exactly similar to those I subsequently observed near the
base of the mountain of Molu.
There are near the main path some large reservoirs in which very
fine fish are found, and the road being led along the banks, or over
the broad dams, it was very picturesque, particularly near the
limestone hills of Piat, where we found a large party of Malays
seeking gold in the quartz which lines the crevices and the caves of
these hills. A very pretty specimen was shown us with the particles
of gold sparkling as if imbedded in crystal.
As it was my companion’s duty to inspect all the stations, we
diverged to the right to visit the antimony mines of Busu. We found
there upwards of fifty Chinese apparently working at the rock on the
steep face of a hill, burrowing here and there in the limestone. The
rocks were very much like those of the mountain of Molu, and
climbing over their sharp surface into little out-of-the-way corners, we
found two or three Chinese scattered here and there, picking out
lumps of antimony from the crevices of the limestone, or perfectly
imbedded in it, and requiring much labour to procure.
These are not really mines—no vein is found, but merely lumps of
ore scattered in every direction.
Just at the foot of the hill in the forest we came upon two Malays
who had just discovered a lump of antimony weighing several tons
which was but a few feet below the surface, and having cleared
away the superincumbent earth, were now covering it over with dry
wood, in order to split the metal, by first raising the temperature by
fire, and then suddenly reducing it by water.
Near Bau the reservoirs increased in number and extent, while
the population became more numerous, and as at each Chinese
house there were several ferocious dogs kept, it was necessary to
be perpetually on one’s guard. The town of Bau was much more
extensive than I could have supposed; I counted above one hundred
shops, and there were many houses besides.
Our attention was particularly drawn to one long, enclosed shed,
filled with Chinese, who evidently, from their appearance and
conduct, were strangers. On arriving at the gold company’s house,
we made inquiries respecting these late arrivals, and the principal
people positively denied any had reached Bau, which was evidently
untrue. These kunsis are regular republics, governed by officers
selected by the multitude: a common workman may suddenly be
elevated to be their leader. They generally choose well, and look
chiefly to the business character of the man put up for their selection.
Though, perhaps, gold-working pays on the whole more than
other labour, the men are kept so very hard at work that the ranks
are not easily filled; yet they are allowed five meals a day, with as
much rice as they can eat, a good supply of salt fish and pork, and
tea always ready. At their meals the Chinese are very fond of
drinking their weak arrack, or samshu, raw, but as nearly
approaching a boiling state as the lips can endure.
Our tour then led us to the neighbourhood of the antimony mines
of Bidi, where the Bornean company are at present working that
metal with success.
The country here is very picturesque: fine open valleys bordered
by almost perpendicular limestone hills, and with an admirable soil.
Occasionally the whole length of a precipice is undermined, forming
extensive open caves, with huge stalactites hanging down at the
extreme edge, giving a beautiful yet fantastic appearance to these
natural dwellings.
We spent a night at a village of the Sau Dayaks, whose long
dwelling was built on a steep hill on the banks of the Sarawak river;
and from thence on nearly to the borders of the Sarawak territory,
was an admirable path constructed by the Chinese to facilitate their
intercourse with Sambas, but unfortunately for travellers, they had
not completed it to Bau.
A four hours’ walk brought us to the last Chinese station, which
was evidently more intended as a resting place for wayfarers than
because its inhabitants were actually much engaged in gold-working
there. We ascended the hills to the village of the Gombang Dayaks,
and heard that a continual stream of small parties of Chinese was
constantly passing within sight of their village. So there was little
doubt that the Chinese population was increasing.
From all the inquiries we made as to the numbers engaged
directly and indirectly in gold-working, we considered there were
nearly three thousand living between the town of Siniawan and the
border. There were about five hundred agriculturists in Suñgei
Tuñgah and its neighbourhood, and perhaps eight hundred in the
town of Kuching, the sago manufactories, and the surrounding
gardens; but these were soon after recruited by the three hundred
fugitives from Sambas, to whom I have before referred; so that the
Chinese population of Sarawak amounted to above four thousand
five hundred before they rose in insurrection, and while seeking to
overthrow the government, ruined themselves.
CHAPTER XIII.
THE CHINESE INSURRECTION.
Secret Societies—Extensive Intercourse—Smuggling—The Gold

Company fined—Punishment of three of its Members—Arrogance
of the Kunsi—A Police Case—Real Causes of the Insurrection—
An Emissary from the Tien Ti Secret Society—Reported
Encouragement given by the Sultan of Sambas—Sambas Nobles
speak Chinese—Their Nurses—The Nobles conspiring—An
Emissary arrives in Brunei—Proposal—Knowledge of the
intended Insurrection—Proposed Attack on the Consulate—The
Tumanggong’s Threat—The Emissary before the Court—Letter
from the Tien Ti Hué—Rumours of intended Insurrection—
Preparations and Inquiries—Commencement of the Revolt—
Useless Warnings—Surprise of the Government House—Danger
of the Rajah—Cowardice of the Chinese—Escape—Swims the
River—Death of Mr. Nicholets—Attacks on the other Houses—On
the Stockade—Gallant Conduct of Mr. Crymble and the Malay
Fortmen—Warm Reception of the Rebels—Death of a Madman—
A Brave Corporal—Escape of Mr. Crymble—His last Blow—
Fortmen again behave well—Confusion in the Town—Peaceful
Assurances—Attempt to organize a Defence—Panic-stricken—
Departure—Conduct of the People—Next Morning—Killed and
Wounded—The Chinese in Power—The Court House—A Check
to Joy—Oath of Fidelity—Courage shown by Abang Fatah—A
Blow struck—Second Descent of the Chinese—A Boat Action—
Gallant Attack—Deaths—Anecdote—The Second Retreat—The
Town in Flames—The Steamer—The Capital recovered—Pursuit
of the Chinese—They retire to the Interior—Attacks of the Land
Dayaks—Foray of the Chinese—Their Fort taken by the Datu
Bandhar—Pursuit—Disorderly Retreat—Critical Position—Brave
Girls—Pass the Frontier—The Men of the Kunsi—A Quarrel—
Stripped of Plunder—Results of the Insurrection—Forces at the
Disposal of the Sarawak Government—Conduct of the People—
Disastrous to the Chinese—New System—Arrival of armed
Chinese from Sambas—Dutch and English Assistance—Revisit
Sarawak—Change—Conduct of the Rajah—Its Effect on the
People—Secret Society at Labuan—Dangers from the Secret
Societies and their Defenders—Curious Incident—Thoughtful
Care of the Rajah.
I shall endeavour to tell the story of the Chinese insurrection
which suddenly broke out in Sarawak in the year 1857, as it appears
to me to be fraught with instruction to us, and if carefully studied,
may be of infinite service to those who have to govern colonies
where the Chinese form a considerable portion of the population.
For many years the Chinese had attempted to form secret
societies in Sarawak; but every effort was made to check their
spread among the people, and it appeared as if success had
attended that policy. To a considerable extent it was the case: but up
in the interior, among the gold workers, the kunsi or company stood
in the place of a secret society, and its members carried on an
extensive intercourse with their fellow-countrymen in Sambas and
Pontianak, and with the Tien Ti secret society in Singapore. I have
described in the last chapter a tour which Mr. Fox and I made among
the settlements of the Chinese in the interior of Sarawak, during
which we became convinced that smuggling was carried on to a
great extent, for, however numerous might be the new immigrants,
the opium revenue did not increase.
At last it was discovered that opium was sent from Singapore to
the Natuna islands, and from thence smuggled into Sarawak and the
Dutch territories; it was traced to the kunsi, which was thereupon
fined 150l.: a very trifling amount, considering the thousands they
had gained by defrauding the revenue, and measures were
immediately taken to suppress the traffic, which, together with the
punishment of three of its members for a gross assault on another
Chinese, were the only grounds of complaint they had against the
Sarawak government.
To show their arrogance, I will enter into the details of this case. A
Chinese woman ran away from her husband, a member of the kunsi,
who followed her to Kuching, and obtained an order from the police
magistrate that she should return with him, but on her refusal, she
was ordered to remain within the stockade. As the case was
peculiar, she was not confined to a cell, but suffered to move about
in the inner court; and some of her friends supplying her with men’s
clothes, she managed to slip out unperceived by the sentry, and
obtained a passage on board a Chinese boat bound for one of the
villages on the coast. Her husband hearing of the place to which she
had removed, followed her with a strong party of the members of the
kunsi, and recovered her. Not satisfied with this, they seized all the
boatmen, and flogged them in the most unmerciful manner, and then
placed them in irons. When let go, they of course brought their
complaint before the police magistrate, and three of the party were
punished for taking the law into their own hands.
These trivial cases were not the real cause of the insurrection, as
the Chinese before that date were greatly excited by the news that
the English had retired from before Canton; and it was of course
added, we had been utterly defeated, and their preparations were
made before the smuggling was discovered, or the members of their
company punished. The secret societies were everywhere in great
excitement, and the Tien Ti sent an emissary over from Malacca and
Singapore, to excite the gold workers to rebellion, and used the
subtle, but false argument, that not only were the English crushed
before Canton, but that the British Government were so discontented
with Sir James Brooke, that they would not interfere if the kunsi only
destroyed him and his officers, and did not meddle with the other
Europeans, or obstruct the trade.
It is also currently reported that the Sambas sultan and his nobles
offered every encouragement to the undertaking, and the Chinese
listened much to their advice, as these nobles can speak to them in
their own language, and are imbued greatly with Chinese ideas. To
explain this state of things, I may mention that they are always
nursed by girls chosen from among the healthiest of the daughters of
the gold workers; and I may add, that about that time there was a
very active intercourse carried on between the Malay nobles of
Sambas and Makota, and that the latter was constantly closeted with
an emissary of the Tien Ti Hué, or secret society, to whom I am
about to refer. It behoves the Dutch authorities to look well to the
proceedings of the native governments within their own territories, as
there is very great discontent, and there is not the slightest doubt
that the nobles are conspiring.
To show that it is not a mere imagination that the Tien Ti secret
society sent emissaries around at that time, I may state that on the
14th of February, four days before the insurrection, a Chinese
named Achang, who had arrived in Brunei from Singapore a few
days previously, and had the year before been expelled from
Sarawak for joining that Hué, came to my house to try and induce
my four Chinese servants to enter it; and added as a sufficient
reason that the kunsi of Sarawak would by that time have killed all
the white men in that country. He also said that he was very
successful in enlisting members among the sago washers and other
labourers in the capital, and that they had made up their minds to
attack my house, and destroy me within a few weeks, and if my
servants did not join the society they would share my fate.
I did not believe what was said about Sarawak, and any warning
of mine would not have reached there for a month, but I did not
altogether neglect this information, which was secretly given me by
my butler, a Chinese, who had lived several years in England, and
whose death by cholera in 1859 I much regretted; but sent to the
sultan and ministers intimation of what I had heard, and the stern
remark of the tumanggong, that if such an attack were made, not a
Chinese should, by the following night, be left alive in the whole
country, effectually curbed them. This Achang, though a very
quarrelsome fellow, had soon a case of just complaint against a
British subject, which he brought before my court; when it was over, I
asked him how he would have obtained a settlement of his claims, if
his intention to murder me had been carried out. I never saw a man’s
countenance change more, and thinking he was about to fall on the
ground, and to clasp my knees, either to beg for pardon, or, what is
more probable, to entreat that I would not believe the story, I told my
writer to lead him out of the court.
At Bau the letter from the Tien Ti Hué was shown to nakodah
Jeludin by the writers of the kunsi, whilst he was detained a prisoner
there, and this was not invented by him as a startling incident, but
mentioned casually in the course of conversation; this Malay
afterwards died fighting bravely in the last charge to break the ranks
of the Chinese.
During the month of November, 1856, rumours were abroad that
the Chinese gold company intended to surprise the stockades, which

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a 10 6. The area of the triangle is half the product of the

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826 Copyright © 2014 Pearson Education, Inc.

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16. Begin by finding A. Use the Law of Sines to find side c.

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a 30 C1  180  B1  A  180  54  37  89

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c1 a In one triangle, the solution is

834 Copyright © 2014 Pearson Education, Inc.

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836 Copyright © 2014 Pearson Education, Inc.

Using a north-south line, the interior angles are found

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51. According to the figure, 54. Using the figure,

Using the figure,

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56. Using the figure, 58. a. Using the figure,

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Let h = the height of the buildings. Using the figure, b  e  800

Using the figure, A  180  150  30

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842 Copyright © 2014 Pearson Education, Inc.

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844 Copyright © 2014 Pearson Education, Inc.

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846 Copyright © 2014 Pearson Education, Inc.

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1 1 33. Use the given radii to determine that

 17(17  16)(17  10)(17  8) c 2  a 2  b2  2ab cos C

 1071  33 11.82  7.32  10.52  2(7.3)(10.5) cos C

 987.1875  31 a 2  b2  c 2  2bc cos A

31. C  180  15  35  130

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43. Assume that Island B is due east of Island A. Let

45. a. Using the figure,

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a b In the figure, b = the guy wire anchored downhill,

a. Using the figure,

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