jmc

Extend $AM$ to $D$ so that $MD=MA$. Then triangles $AMB$ and $DMC$ are congruent, so triangles $ABC$ and $ACD$ have equal area.



The semi-perimeter of triangle $ACD$ is $(7+15+20)/2=21$, so by Heron's formula, the area of triangle $ACD$ is $$\sqrt{21(21-7)(21-15)(21-20)}=\boxed{42}.$$