5. As shown in Figure 1, in the quadrilateral pyramid $P-ABCD$ with a right trapezoid base, $AD$ $/ / BC, \angle ABC=$ $90^{\circ}, PA \perp$ plane $ABCD, PA=3, AD=2, AB=2 \sqrt{3}, BC=6$. Then the size of the dihedral angle $P-BD-A$ is $\qquad$
5.60 .

Draw $A F \perp B D$ at point $F$.
Since $P A \perp$ plane $A B C D$, we know $P F \perp B D$.
Thus, $\angle P F A$ is the size of the dihedral angle $P-B D-A$.
Because $\angle A B C=90^{\circ}, A D / / B C$, so, in

Rt $\triangle A B D$, by $A B=2 \sqrt{3}, A D=2$, we know
$$
B D=4, A F=\sqrt{3} \text {. }
$$

Therefore, $\tan \angle P F A=\frac{P A}{A F}=\frac{3}{\sqrt{3}}=\sqrt{3}$
$$
\Rightarrow \angle P F A=60^{\circ} \text {. }
$$