In triangle ABC,AB = 13, BC = 14, and AC = 15. Let M be the midpoint of BC. Find AM.
Step 1: Since you know the three sides of the triangle, use the Law of Cosines to find the size of angle(B).
Step 2: Now, using angle(B), AB and BM, use the Law of Cosines to find AM.
A
13 15
B 14 C
BM = 7
Using the Law of Cosines we have
AC^2 = BC^2 + AB^2 - 2 (BC * AB) cos (ABC)
15^2 = 14^2 + 13^2 - 2(14 * 13) cos (ABC)
[ 15^2 - 14^2 - 13^2 ] / [ -2 (14*13) ] = cos (ABC)
-140 / -364 = 5/13 = cos (ABC)
So AM can be found as
AM = sqrt [ BM^2 + AB^2 - 2(BM * AB) cos (ABC) ]
AM = sqrt [ 7^2 + 13^2 - 2(7*13) * (5/13) ]
AM = sqrt [ 49 + 169 - 2 * 7 * 5 ]
AM = sqrt [ 49 +169 - 70]
AM = sqrt [ 148 ] = sqrt [ 4 * 37] = 2 sqrt [37] units ≈ 12.166 units