10.055. A circle is inscribed in an isosceles triangle with a vertex angle of $120^{\circ}$ and a lateral side of $a$ (Fig. 10.55). Find the radius of this circle.
Solution.

Given $A B=B C=a, \angle A B C=120^{\circ}, \angle B A C=\angle A C B=30^{\circ}$, $B E=a \sin 30^{\circ}=\frac{a}{2}, S=r \cdot \frac{A B+B C+A C}{2}, A C=2 a \cdot \cos 30^{\circ}=\frac{2 a \sqrt{3}}{2}=a \sqrt{3} ;$ $S=\frac{1}{2} a^{2} \cdot \sin 120^{\circ}=\frac{a^{2} \sqrt{3}}{4} \Rightarrow \frac{a^{2} \sqrt{3}}{4}=\frac{r \cdot(a \sqrt{3}+2 a)}{2} ; r=\frac{a \sqrt{3}(2-\sqrt{3})}{2}$.

Answer: $\frac{a \sqrt{3}(2-\sqrt{3})}{2}$.