SOLUTION: In triangle ABC, {{{ 2a^2 + 4b^2 + c^2 = 4ab + 2ac }}}. Find the value of cosB.

Algebra ->  Trigonometry-basics -> SOLUTION: In triangle ABC, {{{ 2a^2 + 4b^2 + c^2 = 4ab + 2ac }}}. Find the value of cosB.       Log On


   

Question 1177379: In triangle ABC, +2a%5E2+%2B+4b%5E2+%2B+c%5E2+=+4ab+%2B+2ac+. Find the value of cosB.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!

+2a%5E2+%2B+4b%5E2+%2B+c%5E2+=+4ab+%2B+2ac+.
+a%5E2+%2B+a%5E2+%2B4b%5E2+%2B+c%5E2+-+4ab+-+2ac=0+.
a%5E2+-2ac+%2Bc%5E2+%2Ba%5E2+-4ab%2B4b%5E2=0
%28a-c%29%5E2+%2B%28a-2b%29%5E2=0
Here a, b and c are positive real numbers, and if (a - 2b)^2 > 0, then (a - c)^2 is positive so a - 2b = 0 and a - c = 0.
a - 2b = a - c
c = 2b
Plug in the original equation,
2a%5E2+%2B+4b%5E2+%2B+4b%5E2
= 2a^2 + 8b^2
= 4ab + 4ab
= 8ab
0+=+2a%5E2+-+8ab+%2B+8b%5E2
0=+a%5E2+-+4ab+%2B+4b%5E2
= (a - 2b)^2
a - 2b = 0
a = 2b
Plug in original equation,
8b%5E2+%2B+4b%5E2+%2B+4b%5E2+=+4a%28a%2F2%29+%2B+2c%28c%29
= 2a%5E2+%2B+2c%5E2
= 16b%5E2
b%5E2=+%281%2F8%29+a%5E2+%2B+%281%2F8%29+c%5E2
= a%5E2+%2B+c%5E2+-+%28%287%2F8%29+a%5E2+%2B+%287%2F8%29+c%5E2%29
Using Law of Cosines,
2accos(B) = %287%2F8%29+a%5E2+%2B+%287%2F8%29+c%5E2
multiply by 8
16ac *cos(B) = 7a%5E2+%2B+7c%5E2
cos(B) = %287a%5E2+%2B+7c%5E2%29%2F%2816ac%29
=+%287%284b%5E2%29+%2B+7%284b%5E2%29%29%2F%2816%284b%5E2%29%29
= %2856b%5E2%29%2F%2864b%5E2%29
= 56/64
= 7/8
Hence cos(B) = 7/8.