Example 9 As shown in Figure 1.13.7, a billiard table is in the shape of a regular hexagon $A B C D E F$. A ball is struck from the midpoint $P$ of $A B$, hits a point $Q$ on side $B C$, and then bounces off sides $C D, D E, E F, F A$ in sequence, finally hitting a point on $A B$. Let $\angle B P Q=\theta$. Find the range of values for $\theta$.

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Solution 1 Let the points of impact of the ball on each side be $P, Q, R, S, T, U, V$. According to the principle that the angle of incidence equals the angle of reflection, we have $\angle P Q B = \angle R Q C$. Also, $\angle B = \angle C$, so $\triangle Q P B \sim \triangle R Q C$, hence $\frac{B Q}{B P} = \frac{C Q}{C R}$. Similarly, $\frac{B Q}{B P} = \frac{C Q}{C R} = \frac{D S}{D R} = \frac{E S}{E T} = \frac{F U}{F T} = \frac{A U}{A V}$.

Without loss of generality, let the side length of the regular hexagon be $1$, and $B Q = x$. Then $P B = \frac{1}{2}$, $C Q = 1 - x$, $C R = \frac{1 - x}{2 x}$, $D R = \frac{3 x - 1}{2 x}$, $D S = 3 x - 1$, $E S = 2 - 3 x$, $E T = \frac{2 - 3 x}{2 x}$, $F T = \frac{5 x - 2}{2 x}$, $F U = 5 x - 2$, $A U = 3 - 5 x$, $A V = \frac{3 - 5 x}{2 x}$.

Since $Q, R, S, T, U, V$ are all on the sides of the regular hexagon, we get the inequality system (i.e., all expressions take values in $(0,1)$), and solve to get $\frac{3}{7} < x < \frac{3}{5}$.

In $\triangle P B Q$, $B P = \frac{1}{2}$, $\angle P B Q = 120^{\circ}$, $\frac{3}{7} < B Q < \frac{3}{5}$. By the cosine rule, we can find the range of $P Q$ as $\frac{\sqrt{127}}{14} < P Q < \frac{\sqrt{91}}{10}$. By the sine rule, we can find the range of $\angle Q P B$ as
$$
\arcsin \frac{3 \sqrt{3}}{\sqrt{127}} < \theta < \arcsin \frac{3 \sqrt{3}}{\sqrt{91}}
$$

Solution 2 Using the symmetry axis of the figure for reflection, reflecting 5 times as shown in Figure 1.13.7, we can find $B_{1} H = \frac{3 \sqrt{3}}{2}$, $B B_{1} = 3 \sqrt{3}$, and we can also obtain the above results. (Omitted)