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

1 Answer
Mar 28, 2017

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