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
E D
H 144
A 66 B
Consider triangle AHB formed by BE and AD
Angle BAH = 66
Angle AHB =144
But the sum of these > 180 which is impossible
No solution
+1348 
