3. As shown in Figure 3, in isosceles $\triangle ABC$, $AB = AC$, and $P$, $Q$ are points on sides $AC$, $AB$ respectively, such that $AP = PQ = QB = BC$. Then $\angle PCQ =$ $\qquad$
3. $30^{\circ}$.

Let $\angle A=2 \alpha$, then
$$
\begin{array}{l}
\angle A B C=\angle A C B=90^{\circ}-\alpha, \\
\angle B C Q=\angle B Q C=45^{\circ}+\frac{\alpha}{2}, \\
\angle P C Q=\angle A C B-\angle B C Q=45^{\circ}-\frac{3 \alpha}{2} .
\end{array}
$$

From $(A P \cdot 2 \cos 2 \alpha+B Q) 2 \sin \alpha=B C$, we get $2 \sin \alpha(2 \cos 2 \alpha+1)=1$
$$
\begin{array}{l}
\Rightarrow 2 \sin \alpha\left(3-4 \sin ^{2} \alpha\right)=1 \\
\Rightarrow \sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha=\frac{1}{2}
\end{array}
$$
$\Rightarrow \alpha=10^{\circ}$ or $50^{\circ}$ (discard).
Thus, $\angle P C Q=45^{\circ}-\frac{3 \alpha}{2}=30^{\circ}$.