Unit 5.5

1. 5.5
Law of Sines
Copyright © 2011 Pearson, Inc.

2. What you’ll learn about
 Deriving the Law of Sines
 Solving Triangles (AAS, ASA)
 The Ambiguous Case (SSA)
 Applications
… and why
The Law of Sines is a powerful extension of the triangle
congruence theorems of Euclidean geometry.
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3. Law of Sines
In VABC with angles A, B, and C opposite sides
a, b, and c, respectively, the following equation is true:
sin A
a

sin B
b

sinC
c
.
Copyright © 2011 Pearson, Inc. Slide 5.5 - 3

4. Example Solving a Triangle Given
Two Angles and a Side
Solve VABC given that A  38o, B  46o, and a  9.
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5. Example Solving a Triangle Given
Two Angles and a Side
Solve VABC given that A  38o, B  46o, and a  9.
Find C  180o  38o  46o  96o.
Apply the Law of Sines:
sin A
sin B
sin A


a
b
a
sinC
c
sin 38o
9

sin 46o
b
sin 38o
9

sin96o
c
b 
9sin 46o
sin 38o c 
9sin96o
sin 38o
b  10.516 c  14.538
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6. Example Solving a Triangle Given
Two Angles and a Side
Solve VABC given that A  38o, B  46o, and a  9.
The six parts of the triangle are:
A  38o a  9
B  46o b  10.516
C  96o c  14.538
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7. Example Solving a Triangle Given Two Sides
and an Angle (The Ambiguous Case)
Solve VABC given that a  12, c  8, and C  25º .
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8. Example Solving a Triangle Given Two Sides
and an Angle (The Ambiguous Case)
Solve VABC given that a  12, c  8, and C  25º .
B
Use the Law of Sines to find .
sin 30 sin
B

7 6
 
1
 
6sin 30
sin
 
7

 
25.4 or 180 25.4 154.6
If 25.4 , If
Copyright © 2011 Pearson, Inc. Slide 5.5 - 8
B
B B
B
   
 154.6 ,
then 180 30 25.4 124.6 then 180 30 154.6 4.6
7sin124.6
and 11.5 Since this is not possible,
sin 30
B
C C
c

          
 
there is only one triangle.
The six parts of the triangle are:
A  30 , a  7, B  25.4 , b  6, C 124.6 , c 11.5.

9. Example Solving a Triangle Given Two Sides
and an Angle (The Ambiguous Case)
Solve VABC given that a  12, c  8, and C  25º .
Use the Law of Sines to find B. If B is acute:
sin B
sin25º

12
8
sin B 
3sin25º
2
B  sin1 3sin25º
2

 

 
B  39º
Copyright © 2011 Pearson, Inc. Slide 5.5 - 9

10. Example Solving a Triangle Given Two Sides
and an Angle (The Ambiguous Case)
Solve VABC given that a  12, c  8, and C  25º .
Continuing with B acute:
B  39º
A  116º
sin A
sinC

a
c
sin116º
a

sin25º
8
a  17.0
Copyright © 2011 Pearson, Inc. Slide 5.5 - 10

11. Example Solving a Triangle Given Two Sides
and an Angle (The Ambiguous Case)
Solve VABC given that a  12, c  8, and C  25º .
Use the Law of Sines to find B. If B is obtuse:
sin B
sin25º

12
8
sin B 
3sin25º
2
B  180º  sin1 3sin25º
2

 

 
B  141º
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12. Example Solving a Triangle Given Two Sides
and an Angle (The Ambiguous Case)
Solve VABC given that a  12, c  8, and C  25º .
Continuing with B obtuse:
B  141º
A  14º
sin A
sinC

a
c
sin14º
a

sin25º
8
a  4.6
Copyright © 2011 Pearson, Inc. Slide 5.5 - 12

13. Example Solving a Triangle Given Two Sides
and an Angle (The Ambiguous Case)
Solve VABC given that a  12, c  8, and C  25º .
Interpret
One triangle has a  17.0, A  116º , and B  39º .
The other has a  4.6, A  14º , and B  141º .
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14. Example Finding the Height of a Pole
A road slopes 15o above the horizontal, and a vertical
telephone pole stands beside the road. The angle of
elevation of the Sun is 65o, and the pole casts a 15 foot
shadow downhill along the road. Find the height
of the pole.
65º
x
15º
B
A
C
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15. Example Finding the Height of a Pole
65º
x
15º
B
A
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C
Let x  the height of the pole.
BAC  180o  90o  65o  25o
ACB  65o 15o  50o
sin25
sin50

15
x
x 
15sin50
sin25
 27.2
The height of the pole is about 27.2 feet.

16. Quick Review
Given a / b  c / d, solve for the given variable.
1. b
2. c
Evaluate the expression.
3.
8sin 32o
5
Solve for the angle x.
4. sin x  0.2 0o  x  90o
5. sin x  0.2 90o  x  180o
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17. Quick Review Solutions
Given a / b  c / d, solve for the given variable.
1. b
ad
c
2. c
ad
b
Evaluate the expression.
3.
8sin 32o
5
0.848
Solve for the angle x.
4. sin x  0.2 0o  x  90o 11.537o
5. sin x  0.2 90o  x  180o 168.463o
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