Altitudes $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ intersect at $H$. If $\angle BAC = 46^\circ$ and $\angle ABC = 71^\circ$, then what is $\angle AHB$?
Angle ADB = 90°
And angle DBA = 71°
So angle BAH = 90 - 71 = 19°
And angle BEA = 90°
And angle EAB = 46°
So angle ABH = 90 - 46 = 44°
So...in triangle AHB, angle AHB = 180 - 19 - 44 = 117°
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