2. $P$ is a point inside $\triangle A B C$ such that $\angle P B C=$ $30^{\circ}, \angle P B A=8^{\circ}$, and $\angle P A B=\angle P A C=22^{\circ}$. What is $\angle A P C$ in degrees?
2. As shown in Figure 8, extend $A C$ to point $Q$ such that $A B = A Q$. Then
$$
\begin{array}{l} 
\triangle B A P \cong \triangle Q A P . \\
\text { Also, } \angle A P B \\
= 180^{\circ} - \angle P B A - \\
\angle P A B \\
= 150^{\circ}, \\
\text { then } \angle B P Q \\
= 360^{\circ} - \angle A P B - \angle A P Q = 60^{\circ} .
\end{array}
$$

Therefore, $\triangle B P Q$ is an equilateral triangle, and $B C$ is the perpendicular bisector of $P Q$, i.e.,
$$
\angle C P Q = \angle C Q P = \angle P B A = 8^{\circ} .
$$

Thus, $\angle A P C = \angle A P Q - \angle C P Q = 142^{\circ}$.