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Let $A B C$ be an isosceles triangle with $B C$ as its base. Then, $r_1=$
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Verified Answer
The correct answer is:
$R^2 \sin ^2 A$
We have, $A B C$ be an isosceles triangle, $B C$ as its base.
$$
\therefore \quad \angle B=\angle C
$$
We know that,
$$
r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
$$
$$
\begin{aligned}
r_1 & =4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\
\therefore \quad r_1 & =16 R^2 \sin ^2 \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2} \\
& =4 R^2 \sin ^2 \frac{A}{2} \cdot \sin B \sin C \\
& =4 R^2 \sin ^2 \frac{A}{2} \cdot \sin ^2 B \\
& =4 R^2 \sin ^2 \frac{A}{2} \sin ^2\left(\frac{\pi}{2}-\frac{A}{2}\right) \quad[\because \angle B=\angle C] \\
& =4 R^2 \sin ^2 \frac{A}{2} \cos ^2 \frac{A}{2} \\
& =R^2\left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right)^2=R^2 \sin ^2 A
\end{aligned}
$$
$$
\therefore \quad \angle B=\angle C
$$
We know that,
$$
r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
$$
$$
\begin{aligned}
r_1 & =4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\
\therefore \quad r_1 & =16 R^2 \sin ^2 \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2} \\
& =4 R^2 \sin ^2 \frac{A}{2} \cdot \sin B \sin C \\
& =4 R^2 \sin ^2 \frac{A}{2} \cdot \sin ^2 B \\
& =4 R^2 \sin ^2 \frac{A}{2} \sin ^2\left(\frac{\pi}{2}-\frac{A}{2}\right) \quad[\because \angle B=\angle C] \\
& =4 R^2 \sin ^2 \frac{A}{2} \cos ^2 \frac{A}{2} \\
& =R^2\left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right)^2=R^2 \sin ^2 A
\end{aligned}
$$
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