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

Question 1177379: In triangle ABC, . Find the value of cosB.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!

.
.

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^2 + 8b^2
= 4ab + 4ab
= 8ab

= (a - 2b)^2
a - 2b = 0
a = 2b
Plug in original equation,

=
=

=
Using Law of Cosines,
2accos(B) =
multiply by 8
16ac *cos(B) =
cos(B) =
=
=
= 56/64
= 7/8
Hence cos(B) = 7/8.
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