7. Given the dihedral angle $\alpha-l-\beta$ with a plane angle of $60^{\circ}, A, D$ are two points on the line $l$, ray $D B$ is in the plane $\alpha$, and ray $D C$ is in the plane $\beta$. It is known that $\angle B D A=45^{\circ}, \angle C D A=30^{\circ}$, then $\cos \angle B D C$ equals $\qquad$
As shown in the figure, take $B A \perp A D, C A \perp A D$, then
$$
\begin{array}{l}
\angle B A C=60^{\circ} \text{. Let } A D=3 \Rightarrow A B=3, A C=\sqrt{3} \\
\Rightarrow B D=3 \sqrt{2}, C D=2 \sqrt{3}, B C^{2}=12-3 \sqrt{3} .
\end{array}
$$

Thus, in $\triangle B C D$, $\cos \angle B D C$
$$
=\frac{18+12-12+3 \sqrt{3}}{2 \cdot 3 \sqrt{2} \cdot 2 \sqrt{3}}=\frac{2 \sqrt{6}+\sqrt{2}}{8} .
$$