+1671 In right triangle ABC, we have AB=10, BC=22, and angle ABC=90. If M is on AC such that BM is a median of ABC, then what is cos angle ABM? Express your answer in exact form.
+128475 A
10
B 22 C
If BM is a median, then CM = sqrt [ 10^2 + 22^2 ] / 2 = sqrt (584) / 2 = 2sqrt (146) /2 = sqrt (146) = AM
And M will be the circumcenter of the circumradius of triangle ABC....so we have triangle ABM where
AM = BM = sqrt (146) and AB = 10
Using the Law of Cosines
AM ^2 = BM^2 + AB^2 - 2 (BM * AB) cos ( ABM)
146 = 146 + 100 = -2 *sqrt (146)(10) cos (ABM)
-100 / [ -20 sqrt (146 ) ] = cos ABM
cos ABM = 5 /sqrt (146) = (5/146)sqrt (146)
