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Sections 6-1/2
Law of Sines &
Law of Cosines
An oblique triangle is a triangle that has no right angles.
C
a
b
A
c
B
To solve an oblique triangle, you need to know the
measure of at least one side and the measures of any
other two parts of the triangle – two sides, two angles,
or one angle and one side.
2
The following cases are considered when solving oblique triangles.
1. Two angles and any side (AAS or ASA)
A
A
c
c
B
C
2. Two sides and an angle opposite one of them (SSA)
c
C
3. Three sides (SSS)
b
a
c
a
4. Two sides and their included angle (SAS)
c
B
a
3
The first two cases can be solved using the Law of Sines. (The last
two cases can be solved using the Law of Cosines.)
Law of Sines
If ABC is an oblique triangle with sides a, b, and c, then
a  b  c .
sin A sin B sin C
C
C
a
b
h
h
A
c
Acute Triangle
B
a
b
c
A
Obtuse Triangle
B
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Example (ASA):
Find the remaining angle and sides of the triangle.
C
The third angle in the triangle is
A = 180 – A – B
= 180 – 10 – 60
= 110.
10
a = 4.5 ft
4.15 ft b
60
110
A
Use the Law of Sines to find side b and c.
a  c
sin A sin C
4.5  b
sin 110 sin 60
4.5  c
sin 110 sin 10
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
B
0.83 ft
a  b
sin A sin B
b  4.15 feet
c
c  0.83 feet
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Example (SSA):
Use the Law of Sines to solve the triangle.
A = 110, a = 125 inches, b = 100 inches
C
21.26
a  b
sin A sin B
125  100
sin 110 sin B
B  48.74
C  180 – 110 – 48.74
= 21.26
a = 125 in
b = 100 in
110
A
48.74
c
B
48.23 in
a  c
sin A sin C
125 
c
sin 110 sin 21.26
c  48.23 inches
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Example (SSA):
Use the Law of Sines to solve the triangle.
A = 76, a = 18 inches, b = 20 inches
a  b
sin A sin B
18  20
sin 76 sin B
sin B  1.078
C
b = 20 in
a = 18 in
76
B
A
There is no angle whose sine is 1.078.
There is no triangle satisfying the given conditions.
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Example (SSA):
Use the Law of Sines to solve the triangle.
A = 58, a = 11.4 cm, b = 12.8 cm
a  b
sin A sin B
11.4  12.8
sin 58 sin B
B1  72.2
C  180 – 58 – 72.2 = 49.8
C
49.8
b = 12.8 cm
a = 11.4 cm
B1
72.2
58
c
A
10.3 cm
c  b
sin C sin B
c
 12.8
sin 49.8 sin 72.2
c  10.3
Two different triangles can be formed.
Example continues.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
8
Example (SSA) continued:
Use the Law of Sines to solve the second triangle.
A = 58, a = 11.4 cm, b = 12.8 cm
B2  180 – 72.2 = 107.8 
49.8
b = 12.8 cm
a = 11.4 cm
C  180 – 58 – 107.8 = 14.2
c  b
sin C sin B
c
 12.8
sin 14.2 sin 72.2
c  3.3
C
72.2
58
c
A
B1
10.3 cm
C
14.2
b = 12.8 cm
a = 11.4 cm
58
107.8
B2
c
A
3.3 cm
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
9
Area of an Oblique Triangle
Area  1 bc sin A  1 ab sin C  1 ac sin B
2
2
2
C
Example:
Find the area of the triangle.
A = 74, b = 103 inches, c = 58 inches
Area = 1 bc sin A
2
= 1 (103)(58) sin 74
2
 2871.29 sq. inches
103 in
a
b
A
74
c
B
58 in
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The following cases are considered when solving oblique triangles.
1. Two angles and any side (AAS or ASA)
A
A
c
c
B
C
2. Two sides and an angle opposite one of them (SSA)
c
C
3. Three sides (SSS)
b
a
c
a
4. Two sides and their included angle (SAS)
c
B
a
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The last two cases (SSS and SAS) can be solved using the
Law of Cosines.
(The first two cases can be solved using the Law of Sines.)
Law of Cosines
Standard Form
Alternative Form
a  b  c  2bc cos A
2
2
2
b

c

a
cos A 
2bc
b  a  c  2ac cos B
2
2
2
a

c

b
cos B 
2ac
c  a  b  2ab cos C
2
2
2
a

b

c
cos C 
2ab
2
2
2
2
2
2
2
2
2
12
Example:
Find the three angles of the triangle.
2
2
2
a

b

c
cos C 
2ab
2
2
2
8

6

12

2(8)(6)
 64  36  144
96
 44
96
C  117.3
Law of Sines:
C
117.3
8
6
36.3
A
12
 6
sin 117.3 sin B
26.4
B
12
Find the angle
opposite the longest
side first.
B  26.4
A  180 117.3  26.4  36.3
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C
Example:
Solve the triangle.
9.9
Law of Cosines:
b 2  a 2  c 2  2ac cos B
A
67.8
6.2
75
37.2
9.5
B
 (6.2)2  (9.5)2  2(6.2)(9.5) cos 75
 38.44  90.25  (117.8)(0.25882)
 98.20
b  9.9
Law of Sines: 9.9
 6.2
sin 75 sin A
A  37.2
C  180  75  37.2  67.8
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Homework
• WS 13-1
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