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