Question

If in a triangle $ ABC, a^{2}\, \cos^{2}\,A-b^{2}-c^{2}=0,$ then

• A. $\frac{\pi}{4} < A <\frac{\pi}{2}$
• B. $\frac{\pi}{2} < A < \pi$
• C. $A = \frac{\pi}{2} $
• D. $A = \frac{\pi}{4} $

Answer 1

Answer: (B)

We have, $a^{2} \cos ^{2} A-b^{2}-c^{2}=0$
$\Rightarrow a^{2} \cos ^{2} A=b^{2}+c^{2}$
Also in $\triangle A B C$, we have
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{a^{2} \cos ^{2} A-a^{2}}{2 b c}$
$=\frac{-a^{2}\left(1-\cos ^{2} A\right)}{2 b c}=\frac{-a^{2} \sin ^{2} A}{2 b c}<0$
$[\because 0< A< \pi ; a, b, c>0]$
$\Rightarrow \cos A< 0 \Rightarrow $ A lies in IInd quadrant.
$\Rightarrow \frac{\pi}{2}< A< \pi$