jmc

Drawing altitude $\overline{BD}$ splits $△ABC$ into 30-60-90 triangle $ABD$ and 45-45-90 triangle $BCD$:



From 30-60-90 triangle $ABD$, we have $BD=AB/2=10$. From 45-45-90 triangle $BCD$, we have $BC=BD\sqrt{2}=\boxed{10\sqrt{2}}$.