17. (NET 1) Inside an equilateral triangle $A B C$ one constructs points $P$, $Q$ and $R$ such that
$$
\begin{array}{l}
\angle Q A B=\angle P B A=15^{\circ}, \\
\angle R B C=\angle Q C B=20^{\circ}, \\
\angle P C A=\angle R A C=25^{\circ}
\end{array}
$$

Determine the angles of triangle $P Q R$.
17. Let $K, L$, and $M$ be intersections of $C Q$ and $B R, A R$ and $C P$, and $A Q$ and $B P$, respectively. Let $\angle X$ denote the angle of the hexagon $K Q M P L R$ at the vertex $X$, where $X$ is one of the six points. By an elementary calculation of angles we get
$$
\angle K=140^{\circ}, \angle L=130^{\circ}, \angle M=150^{\circ}, \angle P=100^{\circ}, \angle Q=95^{\circ}, \angle R=105^{\circ} \text {. }
$$

Since $\angle K B C=\angle K C B$, it follows that $K$ is on the symmetry line of $A B C$ through A. Analogous statements hold for $L$ and $M$. Let $K_{R}$ and $K_{Q}$ be points symmetric to $K$ with respect to $A R$ and $A Q$, respectively. Since $\angle A K_{Q} Q=\angle A K_{Q} K_{R}=70^{\circ}$ and $\angle A K_{R} R=\angle A K_{R} K_{Q}=70^{\circ}$, it follows that $K_{R}, R, Q$, and $K_{Q}$ are collinear. Hence $\angle Q R K=$ $2 \angle R-180^{\circ}$ and $\angle R Q K=2 \angle Q-$ $180^{\circ}$. We analogously get $\angle P R L=$ $2 \angle R-180^{\circ}, \angle R P L=2 \angle P-$ $180^{\circ}, \angle Q P M=2 \angle P-180^{\circ}$ and $\angle P Q M=2 \angle Q-180^{\circ}$. From these formulas we easily get $\angle R P Q=$ $60^{\circ}, \angle R Q P=75^{\circ}$, and $\angle Q R P=$ $45^{\circ}$.