JEE Advanced 2023 Paper 2 · Q15 · Trigonometric Identities & Equations

Question

PARAGRAPH I (continued) — For the same triangle $ABC$ as in the previous question, the inradius of the triangle $ABC$ is ___.

SolveKar AI · Accepted Answer

Inradius $r = \dfrac{a}{s}$ where $s$ is the semi-perimeter. $s = (AB + BC + CA)/2 = R(\sin A + \sin B + \sin C)$. With $A = \pi/2 - 2C, B = \pi/2 + C$: $\sin A + \sin B + \sin C = \cos 2C + \cos C + \sin C$. Using $\cos 2C = \sqrt{7}/4$ and $\cos C + \sin C = \sqrt{1+\sin 2C} = \sqrt{1+3/4} = \sqrt{7}/2$. So sum = $\sqrt{7}/4 + \sqrt{7}/2 = 3\sqrt{7}/4$. With $R=1$, $s = 3\sqrt{7}/4$. Therefore $r = a/s = (3\sqrt{7}/16)/(3\sqrt{7}/4) = 1/4 = 0.25$.

JEE Advanced 2023 Paper 2 · Q15 · Trigonometric Identities & Equations

Solution