10.214. An isosceles triangle with an acute angle of $30^{\circ}$ at the base is circumscribed around a circle of radius 3. Determine the sides of the triangle.
Solution.

Draw the radius $O E \perp B C$ (Fig. 10.24). Since $\angle O B E=\frac{1}{2} \angle A B C=60^{\circ}$, then $\angle B O E=30^{\circ}$ and $B E=\frac{1}{2} B O$. From $\triangle B E O$ we find $B O^{2}=\frac{1}{4} B O^{2}+9$, from which $B O=2 \sqrt{3}$. In $\triangle A D B$ we have $A B=2 B D$. But $B D=B O+O D=2 \sqrt{3}+3$ and, therefore, $A B=B C=4 \sqrt{3}+6$. Finally, $A C=2 D C=2(B C-B E)=$ $=2(4 \sqrt{3}+6-\sqrt{3})=6 \sqrt{3}+12$.

Answer: $4 \sqrt{3}+6,6 \sqrt{3}+12$.