Unit 2 - Objective 6 - Law of Cosines
Quick Review

We use the law of sines:

  1. when the measure of two sides and the angle opposite one of them is known and
  2. when the measure of two angles and one side of the triangle is known.

There are two other cases when we need to look at:

  1. the measure of two sides and the included angle and
  2. the measure of three sides of a triangle.

We will use the law of cosines to solve these two cases.

The law of cosines
a2 = b2 + c2 - 2bc cos A
b2 = a2 + c2 - 2ac cos B
c2 = a2 + b2 - 2ab cos C


Example 1

Given: b = 14.7 and c = 9.3, A = 46.3°
Solve the triangle.

Solution

We are given:
sidesangles
a = ?A = 46.3°
b = 14.7B = ?
c = 9.3C = ?

Using the law of cosines:
a2= b2 + c2 - 2bc cos A

= (14.7)2 + (9.3)2 - 2 (14.7) (9.3) cos 46.3°

= 216.09 + 86.49 - 273.42 (0.6908824)

= 216.09 + 86.49 - 188.90107

= 113.67893
a 10.7

We can now use the law of cosines or law of sines to find angle B. Using the law of cosines:
b2= a2 + c2 - 2ac cos B
(14.7)2= (10.7)2 + (9.3)2 - 2 (10.7) (9.3) cos B
216.09= 114.49 + 86.49 - 199.02 cos B
15.11= - 199.02 cos B
- 0.075922 cos B
94.3oB

C = 180° - 94.3° - 46.3° = 39.4°


We have:
sidesangles
a = 10.7A = 46.3°
b = 14.7B = 94.3°
c = 9.3C = 39.4°




Example 2

Given: a = 23.31, b = 27.26 and c = 29.17
Solve the triangle.

Solution

We are given:
sidesangles
a = 23.31A = ?
b = 27.26B = ?
c = 29.17C = ?

Using the law of cosines:
a2= b2 + c2 - 2bc cos A
(23.31)2= (27.26)2 + (29.17)2 - 2 (27.26) (29.17) cos A
543.36= 743.11 + 850.89 - 1590.35 cos A
-1050.64= -1590.35 cos A
0.660634 cos A
48.65° A

We can now use the law of cosines or law of sines to find angle B. Using the law of cosines:
b2= a2 + c2 - 2ac cos B
(27.26)2= (23.31)2 + (29.17)2 - 2 (23.31) (29.17) cos B
743.11= 543.36 + 850.89 - 1359.91 cos B
- 651.14= - 1359.91 cos B
0.478811 cos B
61.39°B

C = 180° - B - A

C = 180° - 61.39° - 48.65°

C = 69.96°

We have:
sidesangles
a = 23.31A = 48.65°
b = 27.26B = 61.39°
c = 29.17C = 69.96°






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