In a triangle \[ABC,\,\,2{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}=4ab+2ac,\] then\[\cos B\] is equal to

In a triangle \[ABC,\,\,2{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}=4ab+2ac,\] the

1.	In a triangle \[ABC,\,\,2{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}=4ab+2ac,\] then\[\cos B\] is equal to
A.	0
B.	\[\frac{1}{8}\]
C.	\[\frac{3}{8}\]
D.	\[\frac{7}{8}\]
Answer» E.

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In a triangle \[ABC,\,\,2{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}=4ab+2ac,\] then\[\cos B\] is equal to

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