Example 5 In $\triangle A B C$, $A B=A C, \angle A=80^{\circ}$, $P$ is a point inside $\triangle A B C$, and $\angle P B C=10^{\circ}, \angle P C A=30^{\circ}$. Find the degree measure of $\angle P A C$.
Solution: As shown in Figure 6, construct the symmetric point $P^{\prime}$ of point $P$ with respect to $A C$, and connect $P^{\prime} A$, $P^{\prime} B$, $P^{\prime} C$, and $P^{\prime} P$.

Since $\angle P C A=30^{\circ}$, by symmetry we have
$$
\begin{array}{l}
\angle P C P^{\prime} \\
=2 \angle P C A=60^{\circ}, \\
P C=P^{\prime} C .
\end{array}
$$

Thus, $\triangle P P^{\prime} C$ is an equilateral triangle.
From the given information,
$$
\angle A C B=\frac{1}{2}\left(180^{\circ}-\angle A\right)=50^{\circ} \text {. }
$$

Then $\angle P C B=\angle A C B-\angle P C A=20^{\circ}$,
$$
\begin{array}{l}
\angle B P C=180^{\circ}-(\angle P C B+\angle P B C)=150^{\circ}, \\
\angle B P P^{\prime}=360^{\circ}-\angle B P C-\angle C P P^{\prime} \\
=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ} .
\end{array}
$$

Therefore, $\angle B P C=\angle B P P^{\prime}$.
So, $\triangle P B C \cong \triangle P B P^{\prime}$.
Thus, $\angle P P^{\prime} B=\angle P C B=20^{\circ}$,
$$
\angle P^{\prime} B P=\angle P B C=10^{\circ} \text {. }
$$

Notice that
$$
\begin{array}{l}
\angle C P^{\prime} B=\angle C P^{\prime} P+\angle P P^{\prime} B \\
=60^{\circ}+20^{\circ}=80^{\circ}=\angle C A B .
\end{array}
$$

Therefore, $P^{\prime}, C, B, A$ are concyclic.
Thus, $\angle P^{\prime} A C=\angle P^{\prime} B C=20^{\circ}$.
Hence, $\angle P A C=\angle P^{\prime} A C=20^{\circ}$.
Note: If there is an angle of $30^{\circ}$ in the problem, consider using axial symmetry to restore an equilateral triangle, derive more angle relationships, and achieve the purpose of solving the problem.