In triangle $ABC$, altitudes $AD$, $BE$, and $CF$ intersect at the orthocenter $H$. If $\angle ABC = 49^\circ$ and $\angle ACB = 12^\circ$, then find the measure of $\angle BHC$, in degrees.
See the pic, here :
Since ABC = 49 and ACB = 12
Then BAC = 180 - 49 - 12 = 119
Then we have quadrilateral HFAE
Angle HFA = 90
Angle FAE = Angle BAC = 119
Angle AEH = 90
So angle FHE = Angle BHC = 360 - 90 - 90 - 119 = 61°
