Question

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

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

Answer 1

The correct option is C

isosceles right angled

Explanation for the correct option.

Find the type of triangle.

Using the formulas $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$ and $\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$ the equation $\sin C+\cos C+\sin \left(2B+C\right)-\cos \left(2B+C\right)=2\sqrt{2}$ can be simplified as:

$$\begin{gathered} \left(\sin C+\sin \left(2B+C\right)\right)+\left(\cos C-\cos \left(2B+C\right)\right)=2\sqrt{2} \\ \Rightarrow 2\sin \left(\frac{C+2B+C}{2}\right)\cos \left(\frac{C-2B-C}{2}\right)-2\sin \left(\frac{C+2B+C}{2}\right)\sin \left(\frac{C-2B-C}{2}\right)=2\sqrt{2} \\ \Rightarrow \sin \left(B+C\right)\cos \left(-B\right)-\sin \left(B+C\right)\sin \left(-B\right)=\sqrt{2} \\ \Rightarrow \sin \left(B+C\right)\left[\cos B+\sin B\right]=\sqrt{2} \\ \Rightarrow \sin \left(180°-A\right)\left[\frac{1}{\sqrt{2}}\cos B+\frac{1}{\sqrt{2}}\sin B\right]=1 \\ \Rightarrow \sin A\left[\sin \frac{\mathrm{\pi }}{4}\cos B+\cos \frac{\mathrm{\pi }}{4}\sin B\right]=1 \\ \Rightarrow \sin A\sin \left(\frac{\mathrm{\pi }}{4}+B\right)=1 \end{gathered}$$

Thus $\sin A=1$ and $\sin \left(\frac{\mathrm{\pi }}{4}+B\right)$ and so $A=\frac{\mathrm{\pi }}{2}$ and also

$$\begin{gathered} \frac{\mathrm{\pi }}{4}+B=\frac{\mathrm{\pi }}{2} \\ \Rightarrow B=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Now as $A=\frac{\mathrm{\pi }}{2}$ and $B=\frac{\mathrm{\pi }}{4}$, so $C=\frac{\mathrm{\pi }}{4}$.

Thus $△ABC$ is an isosceles right angled.

Hence, the correct option is C.