In right triangle ABC, we have AB=10, BC=24, 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.
+128408 AC = sqrt (24^2 + 10^2) = sqrt (676) = 26
Let B =(0,0)
Let A =(0,10)
Let C = (24,0)
M = [ (24 + 0)/2 , (10 + 0) /2 ] = ( 12,5)
BM = sqrt [ ( 12 - 0)^2 + ( 5 - 0)^2 ] = sqrt ( 144 + 25) = sqrt (169) = 13
sin BAC = sin BAM = 24 / 26 = 12/13
Using the Law of Sines
sin BAM / BM = sin ABM / AM
(12/13) / 13 = sin ABM / 13
sin (ABM) = 12/13
So
cos ABM = sqrt ( 1 - (12/13)^2 ) = sqrt ( 1 - 144/169) = sqrt (169 -144) / 13 =
sqrt ( 25) /13 =
5 /13
