JEE Advanced 2023 — Paper 2 — Question 35

Let $\mathrm{A}>\mathrm{B}>\mathrm{C}$ $\mathrm{A}-\mathrm{C}=\frac{\pi}{2}$ $a+c=2 b$ $\mathrm{R}=1$ $\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$ $\left(\frac{\pi}{2}+C\right)+B+C=\pi$ $\Rightarrow \mathrm{B}+2 \mathrm{C}=\frac{\pi}{2}$ $2 R \sin A+2 R \sin C=2(2 R \sin B)$ $\sin C+\cos C=2 \cos 2 C$

$$\begin{aligned} & \cos C-\sin C=\frac{1}{2} \\& \sin C=\frac{-1+\sqrt{7}}{4} \text { only } \\& \sin A=\frac{\sqrt{7}+1}{4} \\& \sin B=\frac{\sqrt{7}}{4} \end{aligned}$$

Area of $\triangle A B C=2 R^{2} \sin A \sin B \sin C$ $a=\frac{2}{64}(6 \sqrt{7}) \Rightarrow(64 a)^{2}=1008$