In triangle $ABC,$ $AB = 3,$ $AC = 5,$ $BC = 7,$ and $D$ lies on $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC.$ Find $\cos \angle BAD.$
A
3 5
B D C
7
Law of Cosines
7^2 = 3^2 + 5^2 - 2 (3 * 5) cos (BAC)
cos (BAC ) = [ 7^2 - 3^2 - 5^2 ] / [ -2 * 3 * 5 ] = - 1/2
BAC = 120°
Since BAD is a bisector
BAD = 60°
cos (BAD) = 1 / 2