Altitudes $\overline{AD}$ and $\overline{BE}$ of acute triangle $ABC$ intersect at point $H$. If $\angle AHB = 144^\circ$ and $\angle BAH = 20^\circ$, then what is $\angle ABH$ in degrees?
20 A
E
144 H
B D C
ABH forms a triangle
Angle AHB = 144
Angle BAH = 20
So angle ABH = 180 - 144 - 20 = 16°
+506 
