Example $9 . \angle P O Q=30^{\circ}, A$ is a point on $O Q$, $B$ is a point on $O P$, and $O A=5, O B=12$. Take point $A_{1}$ on $O B$, and take point $A_{2}$ on $A Q$, let $l=A A_{1}+A_{1} A_{2}+A_{2} B$. Find the minimum value of $l$.
Solution: As shown in Figure 13, with the line $O P$ as the axis of symmetry, construct the axis-symmetric figure $\angle P O Q_{0}$ of $\angle P O Q$; with the line $O Q$ as the axis of symmetry, construct the axis-symmetric figure $\angle Q O P_{0}$ of $\angle Q O P$. At this point, the symmetric point of point $A$ about $O P$ is point $A_{0}$ on $O Q_{0}$, and the symmetric point of point $B$ about $O Q$ is point $B_{0}$ on $O P_{0}$. Thus, $O A_{0}=5, O B_{0}=12, \angle A_{0} O B_{0}=90^{\circ}$. By symmetry, we know $A_{0} A_{1}=A A_{1}, B_{0} A_{2}=B A_{2}$. Therefore,
$$
\begin{array}{l}
l=A A_{1}+A_{1} A_{2}+A_{2} B \\
=A_{0} A_{1}+A_{1} A_{2}+A_{2} B_{0} \geqslant A_{0} B_{0} .
\end{array}
$$

Thus, the minimum value of $l$ is the length of $A_{0} B_{0}$.
By the Pythagorean theorem, we get
$$
A_{0} B_{0}^{2}=O A_{0}^{2}+O B_{0}^{2}=5^{2}+12^{2}=169 \text {. }
$$

Therefore, $A_{0} B_{0}=\sqrt{169}=13$.
Hence, the minimum value of $l$ is 13.