Altitudes $\overline{AD}$ and $\overline{BE}$ of acute triangle $ABC$ intersect at point $H$. If $\angle AHB = 144^\circ$ and $\angle BAH = 66^\circ$, then what is $\angle HCA$ in degrees?
A
66
E
144 H
B D C
This is impossible.....triangle AHB would have interior angles that sum to more than 144 + 66 = 210°
+0 
