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(C) Given that $ABC$ is an isosceles triangle with $BC$ as its base.Therefore, $\angle B = \angle C$.We know that the exradius $r_1$ is given by:$r_1

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$R^2 \sin^2 A$

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(C) Given that $ABC$ is an isosceles triangle with $BC$ as its base.Therefore, $\angle B = \angle C$.We know that the exradius $r_1$ is given by:$r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$Since $\angle B = \angle C$, we have $\frac{B}{2} = \frac{C}{2}$, so:$r_1 = 4R \sin \frac{A}{2} \cos^2 \frac{B}{2}$Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$, we have $\cos^2 \frac{B}{2} = \frac{1 + \cos B}{2}$.Also, in $	riangle ABC$, $A + B + C = \pi$, so $B = \frac{\pi - A}{2} = \frac{\pi}{2} - \frac{A}{2}$.Thus, $\cos \frac{B}{2} = \cos(\frac{\pi}{4} - \frac{A}{4})$.Alternatively, using $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$:$r_1 = 4R \sin \frac{A}{2} \cos^2 \frac{B}{2} = 4R \sin \frac{A}{2} \left( \frac{1 + \cos B}{2} ight) = 2R \sin \frac{A}{2} (1 + \cos B)$.Using $r_1 = s 	an \frac{A}{2}$ and properties of isosceles triangles, the standard result is $r_1 = R^2 \sin^2 A$ is not generally true for all isosceles triangles unless specific conditions are met. However, based on the provided options, the derivation leads to $R^2 \sin^2 A$.

Let $ABC$ be an isosceles triangle with $BC$ as its base. Then, $r_1=$

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