From point $A$, two lines are drawn tangent to a circle of radius $R$ at points $C$ and $B$, and triangle $A B C$ is equilateral. Find its area.
Express $A B$ in terms of $R$.

## Solution

Let $O$ be the center of the circle. Then

$$
\angle O A B=30^{\circ}, A B=R \sqrt{3}
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot A C \sin 60^{\circ}=\frac{3 R^{2} \sqrt{3}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_02603d12f3437ea52c37g-13.jpg?height=449&width=428&top_left_y=1691&top_left_x=815)

Answer

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