In $△ A B C$ with usual notations, if $2 a^{2} + 4 b^{2} + c^{2} - 4 a b - 2 a c = 0$ , then $cos A + cos B + cos C$ is equal to:

\frac{1}{4}

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\frac{1}{2}

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\frac{7}{8}

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\frac{11}{8}

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Solution

The correct option is D $\frac{11}{8}$
$2 a^{2} + 4 b^{2} + c^{2} - 4 a b - 2 a c = 0 a^{2} + (2 b)^{2} - 4 a b + c^{2} + a^{2} - 2 a c = 0 (a - 2 b)^{2} + (a - c)^{2} = 0 \Rightarrow a = 2 b = c cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{1}{4}$
Similarily
$cos B = \frac{7}{8}, cos C = \frac{1}{4}$
$cos A + cos B + cos C = \frac{11}{8}$

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