Question

In `triangle ABC`, `a=12, b=8, c=8`, how do you find the cosine of each of the angles?

Answer 1

Use the law of cosines as below

Explanation:

The law of cosines states that `gamma^2=alpha^2+beta^2-2alphabetacosGamma` where `alpha`, `beta`, and `gamma` are the sides of a triangle and `Gamma` is the angle opposite side `gamma` (the angle between sides `alpha` and `beta`.

This law can be rearranged to solve for the cosine of an angle instead of a side.
`cosGamma=(alpha^2+beta^2-gamma^2)/(2alphabeta)`

For each angle, plug in the adjacent sides for `alpha` and `beta` and the opposite side for `gamma`

In your case:

`cosC=(a^2+b^2-c^2)/(2ab)=(12^2+8^2-8^2)/(2*12*8)=3/4`
`cosB=(a^2+c^2-b^2)/(2ac)=(12^2+8^2-8^2)/(2*12*8)=3/4`
`cosA=(b^2+c^2-a^2)/(2bc)=(8^2+8^2-12^2)/(2*8*8)=("-"1)/8`