10.035. From point $A$, two lines are drawn tangent to a circle of radius $R$ at points $B$ and $C$ such that triangle $A B C$ is equilateral. Find its area (Fig. 10.35).
## Solution.

Given $O B=O C=R, A B=A C=B C$. Since $\triangle A B C$ is equilateral, then $S=B C^{2} \frac{\sqrt{3}}{4} ; \angle A B O=90^{\circ}, \angle A B C=60^{\circ}, \angle C B O=30^{\circ}$, $B C=2 \cdot R \cdot \cos 30=\sqrt{3} \cdot R, S=3 R^{2} \frac{\sqrt{3}}{4}$.

Answer: $\quad 3 R^{2} \frac{\sqrt{3}}{4}$.