1. In the right trapezoid $A B C D$, $A D / / B C, A B \perp$ $B C, P$ is a point on $A B$. If $A D=1.5, A B=B C=$ $3, A P=2$, then $\angle P C D=$ $\qquad$
$=1.45^{\circ}$.
As shown in Figure 6, draw $C E \perp$
$A D$ at point $E$, extend $D E$ to point $F$ such that $E F=P B$, and connect $C F$ and $D P$.

It is easy to prove that quadrilateral $A B C E$ is a square and
$\triangle B C P \cong \triangle E C F$.
Thus, $P C=F C, \angle B C P=\angle E C F$.
Therefore, $P D=\sqrt{A P^{2}+A D^{2}}=2.5$.
And $D F=D E+E F=1.5+1=2.5$, so $P D=D F$.
Hence, $\triangle P C D \cong \triangle F C D$.

Therefore, $\angle P C D=\angle F C D=\angle D C E+\angle E C F$
$$
=\angle D C E+\angle B C P=90^{\circ}-\angle P C D \text {. }
$$

Thus, $\angle P C D=45^{\circ}$.