13.6
Apply
the Law of Cosines
Goal · Solve triangles using the law of cosines.
Your
Notes
VOCABULARY
Law of cosines
The law of cosines can be used to solve triangles when two sides and the included angle are known (SAS), or when all these sides are known (SSS).
LAW
OF COSINES
If D ABC has sides of length a, b, and
c as shown, then:
a2 = b2 + c2 - 2bc cos A
b2 = a2 + c2 - 2ac cos B
c2 = _a2 + b2 -2ab cos C_
Your Notes
Example
1
Solve
a triangle for the SAS case
Solve D ABC with a = 7, c
= 12, and B = 26°.
Use the law of cosines to find b.
b2 = a2
+ c2 - 2ac
cos B Law
of cosines
b2 = _72 + 122 - 2(7)(12) cos 26°_ Substitute.
b2 » _42.0_ Simplify.
b » » _6.48_ Take positive
square root.
Use the law of sines to find the
measure of angle A.
When you know all three sides and one angle, you can use
the law of cosines or the law of sines to find the measure of a
second angle.
= Law
of sines
sin A = » _0.4735_ Simplify.
A » sin-1
_0.4735_ »
_28.3°_ Use
inverse sine.
Find the third angle:
C » _180° - 26° - 28.3°_ = _125.7°_.
In DABC, b » _6.48_, A » _28.3°_, and C » _125.7°_.
Checkpoint Solve DABC.
1. b = 15, c = 13, A = 77°
a » 17.5, B » 56.6°, and C » 46.4°
Your Notes
Example 2
Solve a triangle for the SSS case
Solve DABC with a = 7, b =
13, and c = 9.
First find the angle opposite the longest side, AC.
Use the law of cosines to solve for B.
In Example 2, the largest angle is found first to make sure
that the other two angles are acute. This way, when you use the law of
sines to find another angle measure,you will know
that it is between 0°
and 90°.
b2 = a2 + c2
- 2ac cos B
_132_ =
_ 72 + 92 - 2(7)(9) _ cos B
= cos B
_-
0.3095_ » cos B
B » cos-1
_0.3095_ » _108.0°_
Now use the law of sines to find A.
|
|
Law
of sines |
||||||||||||||||
|
sin A = »_0.5121 _
|
Simplify. |
||||||||||||||||
|
A » sin-1 _0.5121_ » _30.8°_ |
Use
inverse sine. |
||||||||||||||||
Find the third angle:
C » _180° - 108.0° - 30.8°_ = _41.2°_.
In DABC, A » _30.8°_,
B » _108.0°_, and C » _41.2°_.
Checkpoint Solve DABC.
2. a = 23, b = 18, c = 20
A » 74.3°, B »
48.9°, and C » 56.8°
Your Notes
HERON'S AREA FORMULA
The area of a triangle with sides
of length a, b, and c is
Area =
Where s = .The variable s is
called the semiperimeter, or half-perimeter, of the triangle.
Example 3
Solve a multi-step problem
Zoo A triangular path around an exhibit at the zoo is shown. Find the
area of the exhibit.
Solution
1. 
Find the semiperimeter s.
s = = =
_143_
2. 
Use Heron's formula to find the area
of DABC.
Area =
=
» _2737_
The area of the exhibit is about _2737_ square feet.
Checkpoint Find the area of D ABC.
3.
about 13.6 square units
4.
about 2170 square units
Homework
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