[ The Law of Sines [ Area of a Triangle (using two sides and the included angle).]

On a line passing through the center $O$ of a circle with radius 12, points $A$ and $B$ are taken such that $O A=15, A B=5$, and $A$ lies between $O$ and $B$. Tangents are drawn from points $A$ and $B$ to the circle, with the points of tangency lying on the same side of the line $O B$. Find the area of triangle $A B C$, where $C$ is the point of intersection of these tangents.
Apply the Law of Sines to triangle $ABC$.

## Solution

Let $M$ and $N$ be the points of tangency of the circle with the lines passing through points $A$ and $B$ respectively,

$\angle O A M=\alpha, \angle O B N=\beta$. Then

$$
\frac{O M}{O A}=\sin \alpha, \frac{O N}{O B}=\sin \beta
$$

Therefore,

$$
\begin{gathered}
\sin \alpha=\frac{4}{5}, \sin \beta=\frac{3}{5}, \cos \alpha=\frac{3}{5}, \cos \beta=\frac{4}{5} \\
B C=\frac{A B \sin \alpha}{\sin (\alpha-\beta)}=\frac{A B \sin \alpha}{\sin \alpha \cos \beta-\cos \alpha \sin \beta}=\frac{100}{7}
\end{gathered}
$$

Thus,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot B C \sin \angle A B C=\frac{1}{2} \cdot 5 \cdot \frac{100}{7} \cdot \frac{3}{5}=\frac{150}{7}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-46.jpg?height=378&width=738&top_left_y=820&top_left_x=660)

## Answer

$\frac{150}{7}$.