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JEE Advance - Mathematics (2023 - Paper 2 Online - No. 14)

Consider an obtuse angled triangle $A B C$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Consider an obtuse angled triangle $A B C$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Consider an obtuse angled triangle $A B C$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
$$ \text { Let } a \text { be the area of the triangle } A B C \text {. Then the value of }(64 a)^2 \text { is } $$ :
Answer
1008

Explanation

Area of $\triangle A B C=\frac{A B \cdot B C \cdot A C}{4 R}$

$$ \begin{aligned} \Rightarrow a & =\frac{8 \sin A \cdot \sin B \sin C}{4} \\\\ & =2 \sin \left(\frac{\pi}{2}-2 C\right) \sin \left(\frac{\pi}{2}+C\right) \sin C \\\\ & =2 \cos 2 C \cdot \cos C \cdot \sin C \\\\ & =\cos 2 C \cdot \sin 2 C \\\\ & =\sqrt{1-\sin ^2 2 C} \cdot \sin 2 C \\\\ & =\sqrt{1-\frac{9}{6}} \cdot \times \frac{3}{4} \\\\ &\Rightarrow a =\frac{3 \sqrt{7}}{16} \\\\ & \therefore (64 a)^2 =1008 \end{aligned} $$

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