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Copyright © 2011 Pearson, Inc.
5.5Law of Sines
Slide 5.5 - 2 Copyright © 2011 Pearson, Inc.
What you’ll learn about
Deriving the Law of Sines Solving Triangles (AAS, ASA) The Ambiguous Case (SSA) Applications
… and whyThe Law of Sines is a powerful extension of the trianglecongruence theorems of Euclidean geometry.
Slide 5.5 - 3 Copyright © 2011 Pearson, Inc.
Law of Sines
In ΔABC with angles A, B, and C opposite sides
a, b, and c, respectively, the following equation is true:
sinAa
=sinB
b=
sinCc
.
Slide 5.5 - 4 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Angles and a Side
Solve ΔABC given that ∠A=38o, ∠B=46o, and a=9.
Slide 5.5 - 5 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Angles and a Side
Find ∠C = 180o−38o−46o =96o.Apply the Law of Sines:sinA
a=
sinBb
sinA
a=
sinCc
sin38o
9=
sin46o
b
sin38o
9=
sin96o
c
b=9sin46o
sin38o c=9sin96o
sin38o
b=10.516 c=14.538
Solve ΔABC given that ∠A=38o, ∠B=46o, and a=9.
Slide 5.5 - 6 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Angles and a Side
The six parts of the triangle are:
∠A=38o a=9
∠B=46o b=10.516
∠C =96o c=14.538
Solve ΔABC given that ∠A=38o, ∠B=46o, and a=9.
Slide 5.5 - 7 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Sides and an Angle (The Ambiguous Case)
Solve ΔABC given that b=12, c=8, and ∠C =25º.
Slide 5.5 - 8 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Sides and an Angle (The Ambiguous Case)
Use the Law of Sines to find ∠B. If ∠B is acute:sinB12
=sin25º
8
sinB=3sin25º
2
B=sin−1 3sin25º2
⎛⎝⎜
⎞⎠⎟
B≈39º
Solve ΔABC given that b=12, c=8, and ∠C =25º.
Slide 5.5 - 9 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Sides and an Angle (The Ambiguous Case)
Continuing with ∠B acute:B≈39º
A≈116º
sinAa
=sinC
csin116º
a=
sin25º8
a≈17.0
Solve ΔABC given that b=12, c=8, and ∠C =25º.
Slide 5.5 - 10 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Sides and an Angle (The Ambiguous Case)
Use the Law of Sines to find ∠B. If ∠B is obtuse:sinB12
=sin25º
8
sinB=3sin25º
2
B=180º−sin−1 3sin25º2
⎛⎝⎜
⎞⎠⎟
B≈141º
Solve ΔABC given that b=12, c=8, and ∠C =25º.
Slide 5.5 - 11 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Sides and an Angle (The Ambiguous Case)
Continuing with ∠B obtuse:B≈141ºA≈14º
sinAa
=sinC
csin14º
a=
sin25º8
a≈4.6
Solve ΔABC given that b=12, c=8, and ∠C =25º.
Slide 5.5 - 12 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle Given Two Sides and an Angle (The Ambiguous Case)
Interpret
One triangle has a ≈17.0, A≈116º, and B≈39º.The other has a≈4.6, A≈14º, and B≈141º.
Solve ΔABC given that b=12, c=8, and ∠C =25º.
Slide 5.5 - 13 Copyright © 2011 Pearson, Inc.
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.
x
15ft15º
65º
B
A
C
Slide 5.5 - 14 Copyright © 2011 Pearson, Inc.
Example Finding the Height of a Pole
x
15ft15º
65º
B
A
C
Let x = the height of the pole.∠BAC =180º−90º−65º=25º
∠ACB=65º−15º=50º
sin25º15
=sin50º
x
x=15sin50ºsin25º
=27.2
The height of the pole is about 27.2 feet.
Slide 5.5 - 15 Copyright © 2011 Pearson, Inc.
Quick Review
Given a / b =c /d, solve for the given variable.1. b2. cEvaluate the expression.
3. 8sin32o
5Solve for the angle x.
4. sinx=0.2 0o < x< 90o
5. sinx=0.2 90o < x<180o
Slide 5.5 - 16 Copyright © 2011 Pearson, Inc.
Quick Review Solutions
Given a / b =c /d, solve for the given variable.
1. b adc
2. c adb
Evaluate the expression.
3. 8sin32o
5 0.848
Solve for the angle x.
4. sinx=0.2 0o < x< 90o 11.537o
5. sinx=0.2 90o < x<180o 168.463o
