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?
C
E D
H
A 20 144 B
ABH forms a triangle
Angle ABH = 180° - 144° - 20° = 16°
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