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 7 D C
Law of Sines
BC^2 = BA^2 + AC^2 - 2 (BA * AC) cos (BAC)
7^2 = 3^2 + 5^2 - 2 (3 * 5) cos (BAC)
[ 7^2 - 3^2 - 5^2 ] / [ -2 (3 * 5) ] = cos (BAC)
-1/2 = cos (BAC)
arccos (-1/2) = BAC = 120°
cos (BAD) = cos (120° / 2) = cos 60° = 1/2
+1768 
