Task B-4.5. On the chord $\overline{A B}$ of the circle $k(S, r)$, two isosceles triangles are constructed on the same side of the chord, with the chord $\overline{A B}$ being their common base. One has its third vertex at the center $S$, and the other at a point $C$ on the circle. If the ratio of their areas is $3:(2 \sqrt{3}+3)$, calculate the measure of the angle between the legs of the triangle $\triangle A B S$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_06_03_ade005c4283eff439485g-22.jpg?height=390&width=357&top_left_y=1559&top_left_x=857)

Image.

(1 point)

Let $a = \overline{AB}$. If the height to the base of triangle $\triangle ABS$ is denoted by $x$, then the height of triangle $\triangle ABC$ is $r + x$.

We have $x = r \cos \frac{a}{2}$.

For the areas of the given triangles, we have

$$
\frac{\frac{(r+x) \cdot a}{2}}{\frac{x \cdot a}{2}} = \frac{2 \sqrt{3} + 3}{3}
$$

After simplifying this expression, we get

$$
\begin{aligned}
\frac{r}{x} + 1 & = \frac{2 \sqrt{3}}{3} + 1 \\
\frac{x}{r} & = \frac{3}{2 \sqrt{3}}
\end{aligned}
$$

From this, we have

$$
\cos \frac{a}{2} = \frac{\sqrt{3}}{2}
$$

Thus, $\alpha = 60^{\circ}$.