In tetrahedron $ABCD$, edge $AB$ has length 3 cm.  The area of face $ABC$ is 15 $\text{cm}^2$ and the area of face $ABD$ is 12 $\text{cm}^2$.  These two faces meet each other at a $30^\circ$ angle.  Find the volume of the tetrahedron in $\text{cm}^3$.
1. **Identify the given information:**
   - Edge \( AB \) has length 3 cm.
   - The area of face \( ABC \) is 15 \(\text{cm}^2\).
   - The area of face \( ABD \) is 12 \(\text{cm}^2\).
   - The angle between faces \( ABC \) and \( ABD \) is \( 30^\circ \).

2. **Calculate the height of face \( ABD \):**
   - The area of \( ABD \) is given by:
     \[
     \text{Area}_{ABD} = \frac{1}{2} \times AB \times h_{ABD}
     \]
     where \( h_{ABD} \) is the height from \( D \) to \( AB \).
   - Plugging in the values:
     \[
     12 = \frac{1}{2} \times 3 \times h_{ABD}
     \]
   - Solving for \( h_{ABD} \):
     \[
     h_{ABD} = \frac{12 \times 2}{3} = 8 \text{ cm}
     \]

3. **Determine the height of the tetrahedron:**
   - The height of the tetrahedron is the perpendicular distance from vertex \( D \) to the plane \( ABC \).
   - Since the faces \( ABD \) and \( ABC \) meet at a \( 30^\circ \) angle, the height of the tetrahedron \( h \) can be found using the sine of the angle:
     \[
     h = h_{ABD} \sin 30^\circ
     \]
   - Given \( \sin 30^\circ = \frac{1}{2} \):
     \[
     h = 8 \times \frac{1}{2} = 4 \text{ cm}
     \]

4. **Calculate the volume of the tetrahedron:**
   - The volume \( V \) of a tetrahedron with base area \( A \) and height \( h \) is given by:
     \[
     V = \frac{1}{3} \times A \times h
     \]
   - Here, the base area \( A \) is the area of face \( ABC \), which is 15 \(\text{cm}^2\), and the height \( h \) is 4 cm:
     \[
     V = \frac{1}{3} \times 15 \times 4 = 20 \text{ cm}^3
     \]

The final answer is \(\boxed{20}\)