1.52. 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.
1.52. Let's draw the radius $O E \perp B C$ (Fig. 1.40). Since $\angle O B E = \frac{1}{2} \angle A B C = 60^{\circ}$, then $\angle B O E = 30^{\circ}$, i.e., $B E = \frac{1}{2} B O$. Then from

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-437.jpg?height=240&width=571&top_left_y=111&top_left_x=307)

Fig. 1.40

$\triangle B E O$ 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: $4 \sqrt{3} + 6, 4 \sqrt{3} + 6$ and $6 \sqrt{3} + 12$.