4. In an isosceles triangle with an angle of $120^{\circ}$, the inscribed circle has a radius of $5 \mathrm{~cm}$. Determine the sides of the triangle.
Solution. Let $O$ be the center of the inscribed circle in the given triangle $ABC$ with an angle of $120^{\circ}$ at vertex $C$. Let $D$ and $E$ be the points of tangency of the circle with $AC$ and $AB$ respectively (see diagram). In the right triangle $ODC, \measuredangle DCO=60^{\circ}$.

From this, it follows that $\overline{DO}=\overline{CO} \sin 60^{\circ}$, i.e., $\overline{CO}=\overline{DO} \frac{2}{\sqrt{3}}=\frac{10}{\sqrt{3}}$. Therefore,

$$
\overline{CE}=\overline{CO}+\overline{OE}=\frac{10}{\sqrt{3}}+5=\frac{5(2+\sqrt{3})}{\sqrt{3}}
$$

From triangle $ACE$ it follows that $\overline{CE}=\overline{AC} \cos 60^{\circ}$, i.e., $\overline{AC}=\frac{10(\sqrt{3}+2)}{\sqrt{3}} \text{ cm}$.

Similarly, it is obtained that $\overline{AE}=\overline{AC} \cos 60^{\circ}$, i.e., $\overline{AB}=10(\sqrt{3}+2) \text{ cm}$.

## II year