A triangle has a regular triangle constructed over one of its sides, a semicircle over another, and a square over the third side. We know that the areas of these three figures are equal. What are the angles of the triangle?

---

Translating the text as requested, keeping the original formatting and line breaks.
Let's denote the sides and angles of the triangle in the usual way and assume that we have constructed a square over side $a$, an equilateral triangle over side $b$, and a semicircle over side $c$.

According to the conditions of the problem,

$$
\begin{gathered}
a^{2}=\frac{b^{2} \cdot \sqrt{3}}{4}=\frac{c^{2} \cdot \pi}{8}, \quad \text { which implies } \\
b=\frac{2 a}{\sqrt[4]{3}}, \quad c=\frac{2 a \sqrt{2}}{\sqrt{\pi}} .
\end{gathered}
$$

By the cosine rule,

$$
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}
$$

Substituting the expressions for $b$ and $c$,

$$
\begin{aligned}
\cos \alpha & =\frac{\frac{4 a^{2}}{\sqrt{3}}+\frac{8 a^{2}}{\pi}-a^{2}}{2 \cdot \frac{2 a}{\sqrt[4]{3}} \cdot \frac{2 a \sqrt{2}}{\sqrt{\pi}}}=0.7950, \quad \text { and thus } \\
\alpha & =37.34^{\circ}
\end{aligned}
$$

Similarly, we find that

$$
\begin{aligned}
\cos \beta & =\frac{a^{2}+\frac{8 a^{2}}{\pi}-\frac{4 a^{2}}{\sqrt{3}}}{2 a \cdot \frac{2 a \sqrt{2}}{\sqrt{\pi}}}=0.3876 \\
\beta & =67.19^{\circ}
\end{aligned}
$$

Finally, $\gamma=180^{\circ}-(\alpha+\beta)=75.47^{\circ}$.