Use the Law of Cosines to solve triangles.

1. 8.2 Law of Cosines
Chapter 8 Applications of Trigonometry

2. Concepts and Objectives
 Law of Cosines
 Use the law of cosines to solve a triangle for a missing
side or angle.
 Use Heron’s formula to find the area of a triangle.

3. Law of Cosines
In any triangle ABC, with sides a, b, and c,
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Instead of trying to learn all three possibilities, I
would suggest learning the pattern. For example:
CB
A
c b
a
  2 2 2
2 cosx y z yz X

4. Law of Cosines
 Example: Solve triangle ABC if A = 42.3°, b = 12.9 m, and
c = 15.4 m.
12.9 m a
C
BA 15.4 m
42.3°

5. Law of Cosines
 Example: Solve triangle ABC if A = 42.3°, b = 12.9 m, and
c = 15.4 m.
12.9 m a
C
BA 15.4 m  2 2 2
2 cosa b c bc A
42.3°
     2 2 2
12.9 15.4 2 12.9 15.4 cos42.3a
2
109.7a
10.47 ma

6. Law of Cosines
 Example: Solve triangle ABC if A = 42.3°, b = 12.9 m, and
c = 15.4 m.
Now we can use a to find B:
12.9 m a
C
BA 15.4 m
2 2 2
2 cosb a c ac B  
42.3°
  2 2 2
12.9 10.47 15.4 2 10.47 15.4 cosB  
166.41 346.86 322.476cosB 
180.45 322.476cosB  
cos .5596B 

7. Law of Cosines
 Example: Solve triangle ABC if A = 42.3°, b = 12.9 m, and
c = 15.4 m.
12.9 m a
C
BA 15.4 m
cos .5596B 
42.3°
 1
cos .5596B 

56.0 
 180 42.3 56.0C   
81.7 

8. Area Formula
 The semiperimeter, s, of a triangle is
and the area of a triangle (Heron’s formula) is:
   
1
2
s a b c
      s s a s b s cA

9. Area Formula
 Example: Find the area of triangle ABC, if a = 82.3 in,
b = 91.7 in, and c = 72.7 in.

10. Area Formula
 Example: Find the area of triangle ABC, if a = 82.3 in,
b = 91.7 in, and c = 72.7 in.
 
1
82.3 91.7 72.7
2
s   
123.35 in
   123.35 123.35 82.3 123.35 91.7 123.35 72.7   A
2
2849.07 in

11. Classwork
 College Algebra
 Page 755: 12-36 (4), page 743: 42-50 (even),
page 606: 44-50 (even)
 Classwork Check 8.2
 Quiz 8.1