9. In tetrahedron $\mathrm{ABCD}$, the length of edge $\mathrm{AB}$ is $3 \mathrm{~cm}$, the area of face $\mathrm{ABC}$ is $15 \mathrm{~cm}^{2}$, and the area of face $\mathrm{ABD}$ is $12 \mathrm{~cm}^{2}$. The angle between these two faces is $30^{\circ}$. Find the volume of the tetrahedron (in $\mathrm{cm}^{3}$).
Let $\mathrm{V}$ be the volume of tetrahedron $\mathrm{ABCD}$, and $\mathrm{h}$ be the height from $\mathrm{D}$ to the base $\mathrm{ABC}$. Then, $\mathrm{V}=\frac{1}{3} \mathrm{hS} \triangle \mathrm{ABC}$. To determine $V$, we only need to determine $h$. Draw $\mathrm{DK} \perp \mathrm{AB}$ at $\mathrm{K}$, and connect $\mathrm{KH}$. By the inverse of the three perpendiculars theorem, $\mathrm{HK} \perp \mathrm{AB}$, thus $\angle \mathrm{DKH}=30^{\circ}$.
From $\mathrm{S} \triangle A B D=\frac{1}{2} \mathrm{DK} \cdot \mathrm{AB}$, we get
$$
\begin{aligned}
\mathrm{DK} & =\frac{2 \mathrm{~S} \triangle A B D}{3}=8 . \\
\therefore \quad \mathrm{h} & =8 \sin 30^{\circ}=4, \\
\mathrm{~V} & =\frac{1}{3} 4 \times 15=20\left(\mathrm{~cm}^{3}\right) .
\end{aligned}
$$