10.280. Determine the angles of an isosceles triangle if its area is related to the area of a square constructed on the base as $\sqrt{3}: 12$.
## Solution.

Let $A C D K$ be the given square (Fig. 10.75), $A B C$ be the given triangle, $A B = B C$, and $B P$ be the height of $\triangle A B C$. Denote $A C = a$, $B P = h$, $S_{\triangle A B C} = \frac{1}{2} a h$, and $S_{A C D K} = a^{2}$. According to the problem, $\frac{\frac{a h}{2}}{a^{2}} = \frac{\sqrt{3}}{12}$ or $\frac{h}{a} = \frac{\sqrt{3}}{6}$. From the right triangle $\triangle A B P: \operatorname{tg} \angle B A P = \frac{B P}{A P} = \frac{h}{\frac{a}{2}} = \frac{\sqrt{3}}{3}$. Therefore, $\angle B A P = \angle B C P = 30^{\circ}$, and $\angle A B C = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

Answer: $30^{\circ}, 30^{\circ}, 120^{\circ}$.