Two faces meet each other

It is obvious that it is an irregular tetrahedron. The volume is given by the formula, $\large\color{#D61F06}V = \dfrac{1}{3}A_{base}h$ . I will consider $\bigtriangleup~ABC$ as the base of the tetrahedron.

Base from my figure and applying Pythagorean Theorem , we have

$A_{ABD} = \dfrac{1}{2}(3)(m)$ $\implies$ $12 = \dfrac{1}{2}(3)(m)$ $\implies$ $\boxed{\color{#69047E}m = 8~cm}$

Solving for $h$ , we have

$sin~30 = \dfrac{h}{m}$ $\implies$ $\boxed{\color{#69047E}h = 4~cm}$

Solving for the volume, we have

$\large V = \dfrac{1}{3}A_{base}h = \dfrac{1}{3}(15)(4) =$ $\boxed{\color{#69047E}\large 20~cm^3}$