10.4 Given a triangle $A B C$. Point $P$ is the incenter. Find the angle $B$, if it is known that $R_{A B C}=R_{A P C}$, where $R_{A B C}, R_{A P C}$ are the radii of the circumcircles of triangles $ABC$ and $APC$ respectively.
10.4. Given a triangle $A B C$. Point $P$ is the incenter. Find the angle $B$, if it is known that $R_{A B C}=R_{A P C}$, where $R_{A B C}, R_{A P C}$ are the radii of the circumcircles of triangles $ABC$ and $APC$ respectively.

Answer: $60^{\circ}$. Hint: Let $\alpha=\angle A B C$. Then $\angle A P C=90^{\circ}+\frac{\alpha}{2}$, since $\angle A P C=180^{\circ}-\frac{1}{2}(\angle B+\angle C)=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\alpha\right)$. Let $R=R_{A B C}=R_{A P C}$. By the known formula (for the Law of Sines) $A C=2 R \sin \alpha$ (in triangle $A B C$) and $A C=2 R \sin \left(90^{\circ}+\frac{\alpha}{2}\right)$ (in triangle $\mathrm{APC})$. From this, we get the equation $\sin \alpha=\sin \left(90^{\circ}+\frac{\alpha}{2}\right) \Leftrightarrow$ $2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}=\cos \frac{\alpha}{2} \Leftrightarrow \sin \frac{\alpha}{2}=\frac{1}{2} \Leftrightarrow \frac{\alpha}{2}=30^{\circ}$ (since $\alpha<180^{\circ}$ ).