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In a triangle ABC, $2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c$ , then the numerical value of $\cos B$ is equal to

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answer is D.

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Detailed Solution

$\begin{aligned} (a - 2 b)^{2} + (a - c)^{2} = 0 \\ a = 2 b; a = c \\ \frac{a}{1} = \frac{b}{1 / 2} = \frac{c}{1} = λ \\ \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} \end{aligned}$

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