# Law of Cosines

How to use the law of cosines to find the side or the angle of the given triangle: formula, 2 examples, and their solutions.

## Formula

### Formula

a2 = b2 + c2 - 2bc cos A

a, b, c: Sides of a triangle
∠A: Angle opposite to side a

Use the law of cosines for the below cases:
2 sides, 1 angle → 1 side
3 sides → 1 angle.

## Example 1

### Solution

Sides: x, 5, 8
Angle opposite to side x: 60º

Then x2 = 52 + 82 - 2⋅5⋅8⋅cos 60º.

To find cos 60º,
draw a 30-60-90 triangle
whose sides are 1, √3, 2.

52 = 25
+82 = +64
-2⋅5⋅8 = -10⋅8 = -80

Find cos 60º.

Cosine is CAH:
Cosine,
Adjacent side (1),
Hypotenuse (2).

So cos 60º = 1/2.

So 52 + 82 - 2⋅5⋅8⋅ cos 60º
= 25 + 64 - 80⋅[1/2].

25 + 64 = 89
-80⋅[1/2] = -40

89 - 40 = 49

x2 = 49

So x = √49 = 7.

x is the length of a side.
So x is plus.

So x = 7.

## Example 2

### Solution

Sides: 6, 5, 4
Angle opposite to side 6: θ

Then 62 = 52 + 42 - 2⋅5⋅4⋅cos θ.

62 = 36

52 = 25
+42 = +16
-2⋅5⋅4 = -10⋅4 = -40

Move 36 to the right side.
And move [-40 cos θ] to the left side.

Then 40 cos θ = 25 + 16 - 36.

+16 - 36 = -20

25 - 20 = 5

40 cos θ = 5

So cos θ = 5/40.

5/40 = 1/8

cos θ = 1/8

So θ = arccos 1/8.

Arccosine: Value

So θ = arccos 1/8.

(arccos 1/8 ≈ 82.82º)