3. In a right triangle $\triangle A B C\left(\measuredangle C=90^{\circ}\right)$, the line passing through the midpoint of the hypotenuse and the center of the inscribed circle intersects the leg $A C$ at point $N$ at an angle of $75^{\circ}$. Determine the acute angles in $\triangle A B C$.
Solution. If $O$ is the center of the inscribed circle, then

$$
\begin{aligned}
& \frac{\alpha+\beta}{2}=45^{\circ} \text { and } \\
& \angle A O B=180^{\circ}-\frac{\alpha+\beta}{2}=135^{\circ} .
\end{aligned}
$$

Due to the above,

$$
\begin{aligned}
& \angle A O N=180^{\circ}-\left(75^{\circ}+\frac{\alpha}{2}\right) \\
& \angle A O M=180^{\circ}-\angle A O N=75^{\circ}+\frac{\alpha}{2} \\
& \angle B O M=135^{\circ}-\angle A O M=60^{\circ}-\frac{\alpha}{2}
\end{aligned}
$$

For $\triangle A O M$, according to the sine theorem,

$$
\frac{\overline{O M}}{\sin \frac{\alpha}{2}}=\frac{\overline{A M}}{\sin \left(75^{\circ}+\frac{\alpha}{2}\right)} \quad \text { i.e. } \quad \overline{A M}=\overline{O M} \frac{\sin \left(75^{\circ}+\frac{\alpha}{2}\right)}{\sin \frac{\alpha}{2}}
$$

For $\triangle B O M$ according to the sine theorem,

$$
\frac{\overline{O M}}{\sin \frac{\beta}{2}}=\frac{\overline{B M}}{\sin \left(60^{\circ}+\frac{\alpha}{2}\right)} \text { i.e. } \quad \overline{B M}=\overline{O M} \frac{\sin \left(60^{\circ}-\frac{\alpha}{2}\right)}{\sin \frac{\beta}{2}} \text {. }
$$

Point $M$ is the midpoint of segment $A B$, so $\overline{A M}=\overline{B M}$, i.e.,

$$
\overline{O M} \frac{\sin \left(75^{\circ}+\frac{\alpha}{2}\right)}{\sin \frac{\alpha}{2}}=\overline{O M} \frac{\sin \left(60^{\circ}-\frac{\alpha}{2}\right)}{\sin \frac{\beta}{2}}
$$

If in the last equality we substitute $\frac{\beta}{2}=45^{\circ}-\frac{\alpha}{2}$, we get

$$
\sin \left(75^{\circ}+\frac{\alpha}{2}\right) \sin \left(45^{\circ}-\frac{\beta}{2}\right)=\sin \left(60^{\circ}-\frac{\alpha}{2}\right) \sin \frac{\alpha}{2}
$$

Using addition theorems, we get

$$
\cos 120^{\circ}-\cos \left(30^{\circ}+\alpha\right)=\cos 60^{\circ}-\cos \left(60^{\circ}-\alpha\right) \Rightarrow
$$

$$
\begin{aligned}
& \cos \left(60^{\circ}-\alpha\right)-\cos \left(30^{\circ}+\alpha\right)=\cos 60^{\circ}-\cos 120^{\circ}=1 \Rightarrow \\
& -2 \sin 45^{\circ} \sin \left(15^{\circ}-\alpha\right)=1 \Rightarrow 2 \sin 45^{\circ} \sin \left(\alpha-15^{\circ}\right)=1 \Rightarrow \\
& \sin \left(\alpha-15^{\circ}\right)=\frac{\sqrt{2}}{2} \Rightarrow \alpha-15^{\circ}=45^{\circ}
\end{aligned}
$$

Thus, the angles of the triangle are $\alpha=60^{\circ}, \beta=30^{\circ}$.