In tetrahedron ABCD, edge AB has length 3 cm. The area of face ABC is  $3c{m}^2$ and the area of the face ABD is  $12c{m}^2.$ These two faces meet each other at  $30°.$  Then the volume of the tetrahedron in  $c{m}^3$  is ________.

Position face ABC on the bottom Since $\left[\Delta ABD\right]=12=\frac{1}{2}.AB.h_{ABD}$  we find  $h_{ABD}=8$
Thus, the height of ABD forms a 30 – 60 – 90 with the height of the tetrahedron. So,  $h=\frac{1}{2}\left(8\right)=4.$  The volume of the tetrahedron is thus $\frac{1}{3}Bh=\frac{1}{3}\mathrm{.3.4}=4$