jmc

Note that triangle $ABC$ is obtuse, so $H$ lies outside triangle $ABC$.



Since triangle $BEC$ is right, $\angle CBE=90^{\circ }-\angle BCE=90^{\circ }-12^{\circ }=78^{\circ }$. Since triangle $BFC$ is right, $\angle BCF=90^{\circ }-\angle CBF=90^{\circ }-49^{\circ }=41^{\circ }$. Therefore, $\angle BHC=180^{\circ }-\angle CBH-\angle BCH=180^{\circ }-78^{\circ }-41^{\circ }=\boxed{61^{\circ }}$.