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?
C
B D
H
144
A 66 B
This is impossible ....the sum of the angles in triangle AHB > 180°
+1234 
