If in a $\triangle ABC, \sin{C}+\cos{C}+\sin{\left(2B+C\right)}-\cos{\left(2B+C\right)}=2\sqrt{2}$, then $\triangle ABC$ is
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equilateral
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isosceles
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right-angled
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obtuse angled
$\because \sin{C}+\cos{C}+\sin{\left(2B+C\right)}-\cos{\left(2B+C\right)}=2\sqrt{2}$
$\Rightarrow \left[\sin{C}+\sin{\left(2B+C\right)}\right]+\left[\cos{C}-\cos{\left(2B+C\right)}\right]=2\sqrt{2}$
$\Rightarrow 2\sin{\left(B+C\right)}\cos{B}+2\sin{\left(B+C\right)}\sin{B}=2\sqrt{2}$
$\Rightarrow \sin{\left(\pi-A\right)}\cos{B}+\sin{\left(\pi-A\right)}\sin{B}=\sqrt{2}$
$\Rightarrow \sin{A}\cos{B}+\sin{A}\sin{B}=\sqrt{2}$
$\Rightarrow \sin{A}\left[\cos{B}+\sin{B}\right]=\sqrt{2}$
Divide both sides by $\sqrt{2}$ we get
$\Rightarrow \sin{A}\left[\dfrac{1}{\sqrt{2}}\cos{B}+\dfrac{1}{\sqrt{2}}\sin{B}\right]=1$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$ we get
$\Rightarrow \sin{A}\left[\sin{\dfrac{\pi}{4}}\cos{B}+\cos{\dfrac{\pi}{4}}\sin{B}\right]=1$
$\Rightarrow \sin{A}\sin{\left(B+\dfrac{\pi}{4}\right)}=1$
$\therefore \sin{A}=1$ and $\sin{\left(B+\dfrac{\pi}{4}\right)}=1$
Hence $A={90}^{0}, \dfrac{\pi}{4}+B=\dfrac{\pi}{2}$
$\Rightarrow B=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$ (on simplification)
$\therefore A={90}^{0}, B={45}^{0}, C={45}^{0}$