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5.5
Law of Sines
Copyright © 2011 Pearson, Inc.
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.
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 2
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 sin B sinC


.
a
b
c
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 3
Example Solving a Triangle Given
Two Angles and a Side
Solve VABC given that A  38o, B  46o, and a  9.
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 4
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 sinC


a
b
a
c
sin 38o sin 46o
sin 38o sin 96o


9
b
9
c
9sin 46o
9sin 96o
b
c
o
sin 38
sin 38o
b  10.516
c  14.538
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 5
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  96
c  14.538
Copyright © 2011 Pearson, Inc.
o
Slide 5.5 - 6
Example Solving a Triangle Given Two Sides
and an Angle (The Ambiguous Case)
Solve VABC given that a  12, c  8, and C  25º.
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 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.
Use the Law of Sines to find B.
sin 30 sin B

7
6
 6sin 30 
B  sin 

 7 
B  25.4
or
If B  25.4 ,
then C  180  30  25.4  124.6
7 sin124.6
and c 
 11.5
sin 30
1
B  180  25.4  154.6
If B  154.6 ,
then C  180  30  154.6  4.6
Since this is not possible,
there is only one triangle.
The six parts of the triangle are:
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 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.
Use the Law of Sines to find B. If B is acute:
sin B sin 25º

12
8
3sin 25º
sin B 
2
1  3sin 25º 
B  sin 


2 
B  39º
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 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.
Continuing with B acute:
B  39º
A  116º
sin A sinC

a
c
sin116º sin 25º

a
8
a  17.0
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 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.
Use the Law of Sines to find B. If B is obtuse:
sin B sin 25º

12
8
3sin 25º
sin B 
2
1  3sin 25º 
B  180º  sin 


2 
B  141º
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 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.
Continuing with B obtuse:
B  141º
A  14º
sin A sin C

a
c
sin14º sin 25º

a
8
a  4.6
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 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.
Interpret
One triangle has a  17.0, A  116º, and B  39º.
The other has a  4.6, A  14º, and B  141º.
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 13
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.
A 65º
x
B
15º
C
Copyright © 2011 Pearson, Inc.
Slide 5.5 - 14
Example Finding the Height of a Pole
A 65º
x
B
15º
C
Copyright © 2011 Pearson, Inc.
Let x  the height of the pole.
BAC  180  90  65  25
o
o
o
o
ACB  65o  15o  50o
sin 25 sin 50

15
x
15sin 50
x
 27.2
sin 25
The height of the pole is about 27.2 feet.
Slide 5.5 - 15