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

The correct option is B $\frac{\pi }{2}<A<\pi$

$a^2{\cos }^{2}A-b^2-c^2=0$
$\Rightarrow {\cos }^{2}A=\frac{b^2+c^2}{a^2}$
We know that ${\cos }^{2}A\le 1$
$\Rightarrow \frac{b^2+c^2}{a^2}\le 1$
$\Rightarrow b^2+c^2\le a^2$
$\Rightarrow b^2+c^2-a^2\le 0$

As $\cos A=\frac{b^2+c^2-a^2}{2bc}$
$\therefore \cos A\le 0$
$\Rightarrow A\in [\frac{\pi }{2},\pi )$

But if $A=\frac{\pi }{2}$, then $b^2+c^2=0$
$\Rightarrow b=c=0$
So, $A=\frac{\pi }{2}$ is not possible.
$\therefore A\in (\frac{\pi }{2},\pi )$