Five, 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 successively strikes sides $C D, D E, E F$, and $F A$, finally hitting a point on side $A B$. Let $\angle B P Q=\theta$, find the range of values for $\theta$.
Hint: Use the principle that the angle of incidence equals the angle of reflection.
Proof 1: Suppose the ball successively hits points $R, S, T, U, V$ on the sides $CD, DE, EF, FA, AB$ (as shown in the figure). According to the principle that the angle of incidence equals the angle of reflection,
$$
\begin{array}{c}
\angle P Q B=\angle R Q C, \\
\angle B=\angle C=120^{\circ}, \\
\text { therefore, } \triangle P Q B \backsim \triangle R Q C,
\end{array}
$$

Thus,
$$
\frac{B Q}{B P}=\frac{C Q}{C R}
$$

Similarly, we can deduce
$$
\begin{aligned}
\frac{B Q}{B P} & =\frac{C Q}{C R}=\frac{D S}{D R}=\frac{E S}{E T} \\
& =\frac{F U}{F T}=\frac{A U}{A V} .
\end{aligned}
$$

Without loss of generality, assume the side length of the regular hexagon is 1, and let the ratio of the above proportion be $2k$, then
$$
\begin{array}{ll}
B Q=k ; & C Q=1-k \\
C R=\frac{1-k}{2 k} ; & D R=\frac{3 k-1}{2 k} ; \\
D S=3 k-1 ; & E S=2-3 k ; \\
E T=\frac{2-3 k}{2 k} ; & F T=\frac{5 k-2}{2 k} ; \\
F U=5 k-2 ; & A U=3-5 k ; \\
A V=\frac{3-5 k}{2 k} . &
\end{array}
$$

For the ball to successively hit each side, $k$ must satisfy the following inequalities:
$$
\left\{\begin{array} { l } 
{ 0 < k < \frac{1}{3} } \\
{ \frac{1}{3} < k < \frac{2}{5} } \\
{ \frac{2}{5} < k < \frac{3}{7} } \\
{ \frac{3}{7} < k < \frac{3}{5} } \\
{ \frac{3}{5} < k < 1 }
\end{array}\right.
$$

The solution to the inequality system is
$$
\frac{3}{7}<k<\frac{3}{5},
$$

which means
$$
\frac{6}{7}<\frac{B Q}{B P}<\frac{6}{5} \text {. }
$$

According to the sine rule,
$$
\frac{B P}{B Q}=\frac{\sin \left(60^{\circ}-\theta\right)}{\sin \theta}=\frac{\sqrt{3}}{2} \cdot \operatorname{ctg} \theta-\frac{1}{2},
$$

which implies
$$
\frac{6}{5}<\frac{\sqrt{3}}{2} \cdot \operatorname{ctg} \theta-\frac{1}{2}<\frac{7}{6} .
$$

Thus,
$$
\frac{3}{10} \sqrt{3}<\operatorname{tg} \theta<\frac{3}{8} \sqrt{3} .
$$

Therefore, the range of $\theta$ is $\operatorname{arctg} \frac{3 \sqrt{3}}{10}<\theta<\operatorname{arctg} \frac{3 \sqrt{3}}{8}$.

Proof 2: As shown in the figure, construct regular hexagons $D C D^{\prime} E^{\prime} F^{\prime} E^{\prime \prime}$, $E^{\prime} F^{\prime} A^{\prime} \cdots F^{\prime \prime}$, and $F^{\prime} A^{\prime} B^{\prime} C^{\prime} \cdots$. Extend $P Q$ to intersect $C D^{\prime}, D^{\prime} E^{\prime}, E^{\prime} F^{\prime}$, $F^{\prime} A^{\prime}, A^{\prime} B^{\prime}$ at points $R^{\prime}, S^{\prime}, T^{\prime}, U^{\prime}, V^{\prime}$, respectively. Suppose the ball hits points $R, S, T, U, V$ on sides $CD, DE, EF, FA, AB$, respectively. Since the angle of incidence equals the angle of reflection, it is easy to prove that
$\triangle Q C R^{\prime} \cong \triangle Q C R, \quad \triangle R^{\prime} D^{\prime} S^{\prime} \cong \triangle R D S$,
$\triangle S^{\prime} E^{\prime} T^{\prime} \cong \triangle S E T, \quad \triangle T^{\prime} E^{\prime} U^{\prime} \cong \triangle T E U$,
$\triangle U^{\prime} A^{\prime} V^{\prime} \cong \triangle U A V$,

The ball hitting point $R$ on $CD$ is equivalent to hitting point $R^{\prime}$ on $C D^{\prime}$; the ball hitting point $S$ on $DE$ is equivalent to hitting point $S^{\prime}$ on $D^{\prime} E^{\prime}$, and so on. Therefore, the broken line $P Q R S T U V$ is transformed into the straight line $P Q R^{\prime} S^{\prime} T^{\prime} U^{\prime} V^{\prime}$. We then discuss the straight line $P V^{\prime}$.

It is easy to prove that $B^{\prime}, A^{\prime}, F^{\prime}, E^{\prime}$ are collinear, and let this line intersect the perpendicular bisector of $A B$ at $M$, with $B^{\prime} M \parallel B A$ and $B^{\prime} M \perp M P$. It is also easy to prove that $F, F^{\prime}, D^{\prime}, B$ are collinear, and $A, C^{\prime}, E^{\prime}, A^{\prime}$ are collinear, because $B^{\prime} A^{\prime} \parallel B A$, so $B B^{\prime} A^{\prime} A$ is a parallelogram: if the extension of $P Q$ can intersect the segments $C D^{\prime}, D^{\prime} E^{\prime}, E^{\prime} F^{\prime}, F^{\prime} A^{\prime}$ within the parallelogram and intersect $A^{\prime} B^{\prime}$, the segment $P V^{\prime}$ must be within the parallelogram $B B^{\prime} A^{\prime} A$. Therefore, it must be that:
$$
\angle P B^{\prime} M<\angle P V^{\prime} M=\angle B P Q=\theta<\angle P A^{\prime} M
$$

Without loss of generality, assume the side length of the regular hexagon is 1, then
$$
\begin{array}{l}
B^{\prime} M=5, \quad A^{\prime} M=4, \quad P M=\frac{3}{2} \sqrt{3} . \\
\operatorname{tg}\left(\angle P B^{\prime} A\right)=\frac{M P}{B^{\prime} M}=\frac{\frac{3}{2} \sqrt{3}}{5}=\frac{3 \sqrt{3}}{10} . \\
\operatorname{tg}\left(\angle P A^{\prime} M\right)=\frac{M P}{A^{\prime} M}=\frac{\frac{3}{2} \sqrt{3}}{4}=\frac{3 \sqrt{3}}{8} . \\
\odot \text { t }, \frac{3 \sqrt{3}}{10}<\operatorname{tg} \theta<\frac{3 \sqrt{3}}{8},
\end{array}
$$

According to $\circledast$,
$$
\frac{3 \sqrt{3}}{10}<\operatorname{tg} \theta<\frac{3 \sqrt{3}}{8},
$$

Therefore, the range of $\theta$ is:
$$
\operatorname{arctg} \frac{3 \sqrt{3}}{10}<\theta<\operatorname{arctg} \frac{3 \sqrt{3}}{8} .
$$