Example 23. A billiard table is in the shape of a regular hexagon ABCDEF. A ball is struck from point P on AB, hits a certain point Q on BC, and then successively strikes the sides CD, DE, EF, FA, and finally hits a certain point V on AB. Let $\angle \mathrm{BPQ} = \theta$. Find the range of values for $\theta$.

Points for improvement: Utilize the principle of equal angles of incidence and reflection.
Solve the following figure, after a series of reflections, transform the broken line $P Q R-$ STJ $\mathrm{Y}$ into a straight line segment $\mathrm{PV}$. Then determine
$$
\angle \mathrm{PB}^{\prime} \mathrm{M}<\angle \mathrm{PV}^{\prime} \mathrm{M}=0<\angle \mathrm{PA}^{\prime} \mathrm{M} \text {. }
$$

Without loss of generality, assume the side length of the hexagon is 1, then
$$
\begin{array}{l}
\mathrm{B}^{\prime} \mathrm{M}=5, \mathrm{~A}^{\prime} \mathrm{M}=4, \mathrm{PM}=\frac{3}{2} \sqrt{3} . \\
\operatorname{tg} \angle \mathrm{PB}^{\prime} \mathrm{A}=\frac{\mathrm{MP}}{\mathrm{B}^{\prime} \mathrm{M}}=\frac{\frac{3}{2} \sqrt{3}}{5}=\frac{3 \sqrt{3}}{10} . \\
\operatorname{tg} \angle \mathrm{PA}^{\prime} \mathrm{M}=\frac{\mathrm{MP}}{\mathrm{A}^{\prime} \mathrm{M}}=\frac{\frac{3}{2} \sqrt{3}}{4}=\frac{3 \sqrt{3}}{8}, \\
\therefore \frac{3 \sqrt{3}}{10}<\operatorname{tg} \theta<-\frac{3 \sqrt{3}}{8} . \\
\therefore \quad \operatorname{arctg} \frac{3 \sqrt{3}}{10}<\theta<\operatorname{arctg} \frac{3 \sqrt{3}}{8} .
\end{array}
$$