In right triangle ABC, we have AB = 10, BC = 20, and angle ABC = 90 degrees. If M is on AC such that BM is a median of triangle ABC, then what is cos ABM?
+129847 A
10 M
B 20 C
AC = sqrt (10^2 + 20^2) = sqrt ( 500) = 10sqrt 5.....so........AM = 5sqrt 5
BM = (10)(20) / (5 sqrt 5) = 40/sqrt 5 = 8 sqrt 5
Using the Law of Cosines
AM^2 = AB^2 + BM^2 -2 (AB * BM) cos ( ABM)
125 = 100 + 320 - 2 ( 10 * 8sqrt 5 ) cos (ABM)
(125 - 100 -320 ) / ( -160sqrt 5) = cos (ABM)
( 59 sqrt 5) / 160 = cos (ABM)
