In triangle $A B C$, angle $B A C$ is $60^{\circ}$, the height dropped from vertex $C$ to side $A B$ is $\sqrt{3}$, and the radius of the circle circumscribed around triangle $A B C$ is 5. Find the sides of triangle $A B C$.
Apply the formula $a=2 R \sin \alpha$.

## Solution

Let $C K$ be the height of triangle $A B C$, and $R=5$ be the radius of the circumscribed circle. Then

$$
\begin{gathered}
A C=\frac{C K}{\sin 60^{\circ}}=2, B C=2 R \sin \angle A=5 \sqrt{3} \\
B K=\sqrt{B C^{2}-C K^{2}}=\sqrt{75-3}=6 \sqrt{2}, A K=A C \cos \angle A=2 \cos 60^{\circ}=1
\end{gathered}
$$

Therefore,

$$
A B=A K+K B=1+6 \sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-23.jpg?height=420&width=560&top_left_y=2002&top_left_x=752)

Answer

$6 \sqrt{2}+1,5 \sqrt{3}, 2$.