Question

In $△ABC$, if $a^2{\cos }^{2}A-b^2-c^2=0$, 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 $\frac{\pi }{2}<A<\pi$

We have,

$a^2{\cos }^{2}A-b^2-c^2=0$
$\to a^2{\cos }^{2}A=b^2+c^2$

Also in $\Delta ABC$, we have
$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{a^2{\cos }^{2}A-a^2}{2bc}$

$=\frac{-a^2(1-{\cos }^{2}A)}{2bc}=\frac{-a^2{\sin }^{2}A}{2bc}<0$ $[\because 0<A<\pi ;a,b,c>0]$

$\Rightarrow$ $\cos A<0$ $\Rightarrow$ A lies in $I{I}^{nd}$ quadrant.
$\Rightarrow \frac{\pi }{2}<A<\pi$