QJEEAdv23Macd06: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Macd06 Solution:  Then the inradius of the triangle  A B C…

Question

Verified Solution • 2025–26 Session moderate

Then the inradius of the triangle A B C is

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Accepted Answer

In radius

r=\frac{\Delta}{S}=\left[\frac{a}{2 R(\sin A+\sin B+\sin C)}\right)

\begin{aligned} r & = \frac{a}{\sin (\frac{π}{2} - 2 C) \sin (\frac{π}{2} + C) + \sin C} \\ = \frac{a}{\cos 2 C + \cos C + \sin C} \\ = \frac{a}{\cos 2 C + \sqrt{1 + \sin 2 C}} \\ = \frac{\frac{3 \sqrt{7}}{16}}{\sqrt{\frac{7}{4}} + \sqrt{\frac{7}{2}}} = \frac{1}{4} \\ \Rightarrow r & = \frac{1}{4} = 0.25 \\ \Rightarrow r & = 0.25 \end{aligned}

QJEEAdv23Macd06: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Macd06 Solution: &nbsp;Then the inradius of the triangle&nbsp; A B C…