In triangle ABC, AB = 13, BC = 14, and AC = 15. Let M be the midpoint of \(\overline{BC}\). Find AM.
+128460 A
13 15
B 14 C
Let B = (0,0) and C = ( 14,0)
So......M = (7,0) ....so BM = 7
Using the Law of Cosines
AC^2 = AB^2 + BC^2 - 2(AB * BC) cos (ABC)
15^2 = 13^2 + 14^2 - 2(13 * 14) cos (ABC)
[ 15^2 - 13^2 - 14^2 ] / [ -2 ( 13 * 14) ] = cos ABC
5/13 = cos ABC
And applying this again we have that
AM = √ [ BM^2 + BA^2 - 2 (BM * BA) cos ABC ]
AM = √ [ 7^2 + 13^2 - 2 ( 7 * 13) (5/13) ]
AM = √[ 49 + 169 - 2*7*5 ]
AM = √ [ 49 + 169 - 70]
AM = √148 = 2√37 units
