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
a2 = b2 + c2 - 2bc cos A
a, b, c: Sides of a triangle
∠A: Angle opposite to side a
Use the law of cosines when:
2 sides, 1 angle → 1 side
3 sides → 1 angle.
Example2 sides, 1 angle → 1 side
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.
Example3 Sides → 1 Angle
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)