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

1 Answer
Mar 28, 2017

Answer:

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#