Question

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

• A. equilateral
• B. isosceles
• C. right-angled
• D. obtuse angled

Answer 1

Answer: (B), (C)

The correct options are
B isosceles
C right-angled

$\because \sin C+\cos C+\sin (2B+C)-\cos (2B+C)=2\sqrt{2}$
$\Rightarrow [\sin C+\sin (2B+C)]+[\cos C-\cos (2B+C)]=2\sqrt{2}$
$\Rightarrow 2\sin (B+C)\cos B+2\sin (B+C)\sin B=2\sqrt{2}$
$\Rightarrow \sin (\pi -A)\cos B+\sin (\pi -A)\sin B=\sqrt{2}$
$\Rightarrow \sin A\cos B+\sin A\sin B=\sqrt{2}$
$\Rightarrow \sin A[\cos B+\sin B]=\sqrt{2}$
Divide both sides by $\sqrt{2}$ we get
$\Rightarrow \sin A[\frac{1}{\sqrt{2}}\cos B+\frac{1}{\sqrt{2}}\sin B]=1$
We know that $\sin \frac{\pi }{4}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ we get
$\Rightarrow \sin A[\sin \frac{\pi }{4}\cos B+\cos \frac{\pi }{4}\sin B]=1$
$\Rightarrow \sin A\sin (B+\frac{\pi }{4})=1$
$\therefore \sin A=1$ and $\sin (B+\frac{\pi }{4})=1$
Hence $A={90}^0,\frac{\pi }{4}+B=\frac{\pi }{2}$
$\Rightarrow B=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$ (on simplification)
$\therefore A={90}^0,B={45}^0,C={45}^0$