JEE Advanced 2023 Paper 2 · Q14 · Trigonometric Identities & Equations

Question

PARAGRAPH I — Consider an obtuse-angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\pi/2$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius $1$. Let $a$ be the area of the triangle $ABC$. Then the value of $(64a)^2$ is ___.

SolveKar AI · Accepted Answer

Sides $AB, BC, CA$ in AP. Let $B$ be largest angle, $C$ smallest, with $B - C = \pi/2$. Using $A + B + C = \pi$: $A = \pi/2 - 2C$. Sides in AP: $2BC = AC + AB$ → $4R\sin A = 2R\sin B + 2R\sin C$ → $2\sin A = \sin B + \sin C$. Substituting $A = \pi/2 - 2C$, $B = \pi/2 + C$: $2\cos 2C = \cos C + \sin C$. This simplifies to $\cos C - \sin C = 1/2 \Rightarrow \sin 2C = 3/4$. Area $a = \dfrac{abc}{4R} = \dfrac{8\sin A\sin B\sin C}{4} = 2\cos 2C\cos C\sin C = \cos 2C\sin 2C$. With $\sin 2C = 3/4$, $\cos 2C = \sqrt{1-9/16} = \sqrt{7}/4$. Area $= 3\sqrt{7}/16$. $64a = 12\sqrt{7}$, so $(64a)^2 = 144\cdot 7 = 1008$.

JEE Advanced 2023 Paper 2 · Q14 · Trigonometric Identities & Equations

Solution