In a traingle $A B C, 2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c,$ then numerical value of $cos B$ is equal to

0

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\frac{1}{8}

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\frac{3}{8}

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\frac{7}{8}

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Solution

The correct option is D $\frac{7}{8}$
$2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c \Rightarrow (a - 2 b)^{2} + (a - c)^{2} = 0 \Rightarrow a = 2 b and a = c \Rightarrow a = 2 b = c or \frac{a}{1} = \frac{b}{\frac{1}{2}} = \frac{c}{1}$
so that $cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a} = \frac{1 + 1 - (\frac{1}{4})}{2 \times 1 \times 1} = \frac{7}{8}$

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