+501 In right triangle $ABC$, we have $AB = 10$, $BC = 24$, and $\angle ABC = 90^\circ$. If $M$ is on $\overline{AC}$ such that $\overline{BM}$ is a median of $\triangle ABC$, then what is $\cos \angle ABM$?
+129852 AC = sqrt ( 24^2 + 10^2) = sqrt [ 576 + 100 ] = sqrt (676) = 26
So.... AM = 13
And a median drawn from the hypotenuse to the opposite side has (1/2) length of the hypotenuse
So....BM = 13
By the Law of Cosines
AM^2 = AB^2 + BM^2 - 2 ( AB * BM) cos ABM
13^2 = 10^2 + 13^2 - 2 ( 10 *13) cos ABM
169 = 100 + 169 - 260 cos ABM
-100 / -260 = cos ABM
5 / 13 = cos ABM
