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In triangle $A B C \angle B=110^{\circ}, \angle C=50^{\circ}$. On side $A B$ a point $P$ is chosen such that $\angle P C B=30^{\circ}$, and on side $A C$ a point $Q$ is chosen such that

$\angle A B Q=40^{\circ}$. Find the angle $Q P C$.
From the condition, it follows that $\angle B P C=40^{\circ}, \angle Q B C=70^{\circ}$.

First method. Draw a ray symmetric to ray $C P$ with respect to line $A C$ (see the left figure). Let $M$ be the point of intersection of this ray with ray $B Q$. Since $\angle P C M=40^{\circ}=\angle P B M$, quadrilateral $P B C M$ is cyclic, so $\angle M P C=\angle M B C=70^{\circ}$. Therefore, $\angle P M C=180^{\circ}-70^{\circ}-40^{\circ}=70^{\circ}$, which means triangle $M R C$ is isosceles. Its bisector $C A$ is the perpendicular bisector of side $P M$, so points $M$ and $P$ are symmetric with respect to it.

Therefore, $\angle Q P C=\angle Q M C=\angle B P C=40^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_5ff42d6d57f9dccbb1f5g-15.jpg?height=402&width=2026&top_left_y=91&top_left_x=13)

Second method. Reflect point $B$ with respect to line $A C$, then $\angle A B^{\prime} C=\angle B=110^{\circ}$ (see the center figure). Since $\angle A=180^{\circ}-\left(50^{\circ}+110^{\circ}\right)=20^{\circ}=\angle P C A$, we have $P A=P C$. Moreover, $\angle A P C=140^{\circ}=2\left(180^{\circ}-\angle A B^{\prime} C\right)$.

Therefore, $P$ is the center of the circumcircle of triangle $A B^{\prime} C$. Thus, $\angle P B^{\prime} A=\angle P A B^{\prime}=40^{\circ}$. Since $\angle Q B^{\prime} A=\angle Q B A=40^{\circ}$, points $P, Q$, and $B^{\prime}$ lie on the same line. Therefore, $\angle Q P C=\angle B^{\prime} P C=2 \angle B^{\prime} A C=40^{\circ}$.

Third method. Let $O$ be the center of the circumcircle of triangle $B C Q$. Then $\angle O C Q=90^{\circ}-\angle Q B C=20^{\circ}=\angle P C Q$, so $O$ lies on ray $C P$ (see the right figure). Moreover, triangle $B C Q$ is acute, so point $O$ lies inside it.

$\angle B Q O=90^{\circ}-\angle B C Q=40^{\circ}=\angle B P O$. Therefore, $P B O Q$ is a cyclic quadrilateral. Thus, $\angle Q P C=\angle Q B O=\angle O Q B=40^{\circ}$.

## Answer

$40^{\circ}$.