2. The lines drawn to the legs of a right triangle form an angle $\varphi$. If it is known that $\operatorname{tg} \varphi=\frac{3}{4}$, find the angles of the triangle.
Solution. Let $\triangle A B C$ be a right-angled triangle at vertex $C$, with legs $\overline{A C}=b, \overline{B C}=a$ and hypotenuse $\overline{A B}=c$. Let $T$ be the centroid of the triangle, $A_{1}$ and $B_{1}$ be the midpoints of sides $B C$ and $A C$, respectively, and the lengths of the medians are $\overline{A A_{1}}=t_{a}$ and $\overline{B B_{1}}=t_{b}$. From $\cos ^{2} \varphi=\frac{1}{1+\operatorname{tg}^{2} \varphi}$, it follows that $\cos \varphi=\frac{4}{5}$. If we use the fact that $\measuredangle A T B_{1}=\measuredangle A_{1} T B=\varphi$ and $\measuredangle A T B=180^{\circ}-\varphi$, and apply the cosine rule in triangles $T A_{1} B, T B_{1} A$ and $T A B$, we get

$$
\begin{aligned}
& \frac{a^{2}}{4}=\frac{t_{a}^{2}}{9}+\frac{4 t_{b}^{2}}{9}-\frac{4 t_{a} t_{b} \cos \varphi}{9} \\
& \frac{b^{2}}{4}=\frac{t_{b}^{2}}{9}+\frac{4 t_{a}^{2}}{9}-\frac{4 t_{a} t_{b} \cos \varphi}{9} \\
& c^{2}=\frac{4 t_{a}^{2}}{9}+\frac{4 t_{b}^{2}}{9}+\frac{8 t_{a} t_{b} \cos \varphi}{9}
\end{aligned}
$$

From the Pythagorean theorem and the above equations, by simplifying the expressions, we get

$$
\frac{20 t_{a}^{2}}{9}+\frac{20 t_{b}^{2}}{9}-\frac{32 t_{a} t_{b} \cos \varphi}{9}=\frac{4 t_{a}^{2}}{9}+\frac{4 t_{b}^{2}}{9}+\frac{8 t_{a} t_{b} \cos \varphi}{9}
$$

If we substitute $\cos \varphi=\frac{4}{5}$ into the last equation, we get

$$
\frac{16}{9}\left(t_{a}-t_{b}\right)^{2}=0
$$

so $t_{a}=t_{b}$, meaning the triangle is an isosceles right-angled triangle. Therefore, the angles of the triangle are $90^{\circ}, 45^{\circ}, 45^{\circ}$.