In a $△ A B C, 2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c$ , then $cos B$ 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 $\frac{7}{8}$
Given, $2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c$
$\Rightarrow a^{2} + (2 b)^{2} - 4 a b + a^{2} + c^{2} - 2 a c = 0$
$\Rightarrow (a - 2 b)^{2} + (a - c)^{2} = 0$
$\Rightarrow a = 2 b = c$
Now, $c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$
$= \frac{c^{2} + c^{2} - {(\frac{c}{2})}^{2}}{2 \times c \times c} = \frac{2 c^{2} - \frac{c^{2}}{4}}{2 c^{2}}$
$\Rightarrow cos B = \frac{7}{8}$ .

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