Faces $ABC$ and $BCD$ of tetrahedron $ABCD$ meet at an angle of $30^\circ$. The area of face $ABC$ is $120$, the area of face $BCD$ is $80$, and $BC=10$. Find the volume of the tetrahedron.
1. **Identify the given information:**
   - The angle between faces $ABC$ and $BCD$ is $30^\circ$.
   - The area of face $ABC$ is $120$.
   - The area of face $BCD$ is $80$.
   - The length of edge $BC$ is $10$.

2. **Calculate the height of triangle $BCD$ relative to side $BC$:**
   - The area of triangle $BCD$ is given by:
     \[
     \text{Area}_{BCD} = \frac{1}{2} \times BC \times \text{height}_{BCD}
     \]
   - Plugging in the known values:
     \[
     80 = \frac{1}{2} \times 10 \times \text{height}_{BCD}
     \]
   - Solving for $\text{height}_{BCD}$:
     \[
     \text{height}_{BCD} = \frac{80 \times 2}{10} = 16
     \]

3. **Determine the altitude of the tetrahedron to face $ABC$:**
   - The height of the tetrahedron from vertex $D$ perpendicular to face $ABC$ can be found using the sine of the angle between the faces $ABC$ and $BCD$.
   - Given that $\sin 30^\circ = \frac{1}{2}$:
     \[
     h_{ABC} = 16 \sin 30^\circ = 16 \times \frac{1}{2} = 8
     \]

4. **Calculate the volume of the tetrahedron:**
   - The volume $V$ of a tetrahedron is given by:
     \[
     V = \frac{1}{3} \times \text{base area} \times \text{height}
     \]
   - Here, the base area is the area of face $ABC$, which is $120$, and the height is $h_{ABC}$, which is $8$:
     \[
     V = \frac{1}{3} \times 120 \times 8 = \frac{1}{3} \times 960 = 320
     \]

The final answer is $\boxed{320}$