jmc

It is clear that $DX=8$ and $CX=10$ where $X$ is the foot of the perpendicular from $D$ and $C$ to side $AB$. Thus $[DXC]=\frac{ab\sin c}{2}=20=5\cdot h\to h=4$ where h is the height of the tetrahedron from $D$. Hence, the volume of the tetrahedron is $\frac{bh}{3}=15\cdot \frac{4}{3}=\boxed{20}$.