In [tetrahedron](https://artofproblemsolving.com/wiki/index.php/Tetrahedron) $ABCD$, [edge](https://artofproblemsolving.com/wiki/index.php/Edge) $AB$ has length 3 cm. The area of [face](https://artofproblemsolving.com/wiki/index.php/Face) $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$. These two faces meet each other at a $30^\circ$ angle. Find the [volume](https://artofproblemsolving.com/wiki/index.php/Volume) of the tetrahedron in $\mbox{cm}^3$.
Position face $ABC$ on the bottom. Since $[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}$, we find that $h_{ABD} = 8$. Because the problem does not specify, we may assume both $ABC$ and $ABD$ to be isosceles triangles. Thus, the height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron. So, $h = \frac{1}{2} (8) = 4$. The volume of the tetrahedron is thus $\frac{1}{3}Bh = \frac{1}{3} \cdot15 \cdot 4 = \boxed{020}$.