Question

If in a $△ABC,$ $\sin C+\cos C+\sin (2B+C)-\cos (2B+C)=2\sqrt{2},$ then $△ABC$ is

• A. isosceles
• B. equilateral
• C. right-angled isosceles
• D. right-angled but not isosceles

Answer 1

The correct option is C right-angled isosceles

$\sin C+\cos C+\sin (2B+C)-\cos (2B+C)=2\sqrt{2}$
$\Rightarrow \sin C+\sin (2B+C)+\cos C-\cos (2B+C)=2\sqrt{2}$
$\Rightarrow 2\sin (B+C)\cos B+2\sin B\sin (B+C)=2\sqrt{2}$
$\Rightarrow 2\sin (\pi -A)[\cos B+\sin B]=2\sqrt{2}[\because A+B+C=\pi ]$

$\Rightarrow \sin A\left[\sqrt{2}(\sin B\cdot \frac{1}{\sqrt{2}}+\cos B\cdot \frac{1}{\sqrt{2}})\right]=\sqrt{2}$
$\Rightarrow \sin A\cdot \sin (B+\frac{\pi }{4})=1$
It is possible only if $\sin A=1$ and $\sin (B+\frac{\pi }{4})=1$
So, $A=\frac{\pi }{2}$ and $B+\frac{\pi }{4}=\frac{\pi }{2}$
$\Rightarrow A=\frac{\pi }{2},B=C=\frac{\pi }{4}$