If in a triangle ABC, $a^{2} {Cos}^{2} A = b^{2} + c^{2}, then$

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0 < $A < \frac{π}{4}$

$\frac{π}{4} < A < \frac{π}{2}$

$\frac{π}{2} < A < π$

$A = \frac{π}{2}$

answer is C.

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Detailed Solution

CosA = $\frac{b^{2} + c^{2} - a^{2}}{2 bc} = \frac{a^{2} {Cos}^{2} A - a^{2}}{2 bc} = \frac{- a^{2} {Sin}^{2} A}{2 bc} < 0$
$\Rightarrow \frac{π}{2} < A < π$

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