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
BD^2 = AB^2 + AC^2 - 2(AB * AC) cos BAC
7^2 = 3^2 + 5^2 - 2 (3 * 5) cos BAC
cos BAC = [ 7^2 - 3^2 - 5^2 ] / [ -2 (3 * 5) ] = -1/2
arccos (-1/2) = measure of BAC = 120°
Since AD is an angle bisector
BAD = 60°
cos (BAD) = cos (60°) = 1/2
+1858 
