3. $P$ is a point on the extension of the diameter $A B$ of $\odot O$, $P C$ is tangent to $\odot O$ at point $C$, and the angle bisector of $\angle A P C$ intersects $A C$ at point $Q$. Then $\angle P Q C=$ $\qquad$
$3.45^{\circ}$.
As shown in Figure 4, connect $O C$. Since $P C$ is tangent to $\odot O$ at point $C$, then $O C$ $\perp P C$.

Since $O A=$ $O C$, therefore,
$$
\begin{array}{l}
\angle O A C=\angle O C A=\frac{1}{2} \angle P O C . \\
\text { Also } \angle A P Q=\angle C P Q=\frac{1}{2} \angle A P C \text {, so } \\
\angle P Q C=\angle P A Q+\angle A P Q \\
=\frac{1}{2}(\angle P O C+\angle A P C)=\frac{1}{2} \times 90^{\circ}=45^{\circ} .
\end{array}
$$