3. As shown in the figure,
$A B C D$ is a square,
$E$ is the midpoint of $A B$. If
$\triangle D A E$ and
$\triangle C B E$ are folded along the dotted lines $D E$ and $C E$ respectively, so that $A E$ and $B E$ coincide, and the point where $A$ and $B$ coincide is denoted as $P$, then the dihedral angle between plane $P C D$ and plane $E C D$ is $\qquad$ degrees.
3. $30^{\circ}$.

As shown in the figure, draw $P F \perp C D$, and connect $E F$. Since $F$ is the midpoint of $C D$, we know that $E F \perp C D$. Therefore, $\angle P F E$ is the plane angle of the dihedral angle formed by the planes $P C D$ and $E C D$.
Let the side length of the square $A B C D$ be $a$. Then
$$
\begin{aligned}
P E & =D F=\frac{1}{2} a, \\
P D & =P C=E F=a, \\
P F & =\sqrt{P D^{2}-D F^{2}} \\
& =\sqrt{a^{2}-\left(\frac{1}{2} a\right)^{2}}=\frac{\sqrt{3}}{2} a .
\end{aligned}
$$

Thus, $\cos \angle P F E=\frac{P F^{2}+E F^{2}-P E^{2}}{2 \cdot P F \cdot E F}=\frac{\sqrt{3}}{2}$.