7. The faces $ABC$ and $BCD$ of tetrahedron $ABCD$ form a $30^{\circ}$ angle, the area of $\triangle ABC$ is 120, the area of $\triangle BCD$ is 80, and $BC=10$. Find the volume of this tetrahedron.
$$
\begin{array}{c}
\text { 7. Through } D \text { draw } D H \\
\perp B C \text { at } H, \\
\because \quad B C=10, \\
S_{\triangle B C D}=80, \\
\therefore \quad D H=16 .
\end{array}
$$

Through $D$ draw a perpendicular to plane $A B C$ intersecting it at: $P$, then $\triangle H P D$ is a right triangle, $P H$ is the projection of $D H$ onto plane $A B C$, by the theorem of three perpendiculars, $B C \perp P H$, so $\angle D H P$ is the dihedral angle between plane $A B C$ and plane $B D C$, hence $\angle D H P = 30^{\circ}$. Therefore, $D P=\frac{1}{2} D H=8$.
Thus, the volume of the tetrahedron is
$$
\frac{1}{3} D P \cdot S_{\triangle B C D}=\frac{1}{3} \times 8 \times 80=213.33
$$