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)
[7^2 - 3^2 - 5^2 ] / [ -2 * 3 *5] = cos (BAC) = -1/2
arccos (-1/2) = BAC = 120°
So cos (BAD) = cos (120 / 2) = cos 60 = 1/2
+136 
