Posted by Kalia on Saturday, July 23, 2011 at 3:59am.
Trigonometry problem:
The sides AB and AC of triangle ABC are of lengths 4 and 7 respectively. M is the midpoint of BC. AM is of length 3.5, Find the length of BC.

Math  Reiny, Saturday, July 23, 2011 at 7:12am
Complete the parallelogram ABPC , where BP=7 and PC=4
In a gram , diagonals bisect each other, so M is the midpoint of both diagonals and AP = 2(3.5) = 7
(looks like we have isosceles triangles, but that doesn't matter)
let angle PAC =Ø
let MC = a , thus BC = 2a for later
In triangle APC, by cosine law
4^2 = 7^2 + 7^2  2(7)(7)cosØ
cosØ = 82/98 = 41/49
in triangle AMC
a^2 = 3.5^2 + 7^2  2(3.5)(7)cosØ
= 61.25  49(41/49) = 20.25
a=√20.25 = 4.5
So BC = 2a = 2(4.5) = 9
better check my arithmetic, I have made some silly errors lately. 
Math  Kalia, Saturday, July 23, 2011 at 9:22am
Thank You!