+23 Altitudes $\overline{AX}$ and $\overline{BY}$ of acute triangle $ABC$ intersect at $H$. If $\angle BAC = 43^\circ$ and $\angle ABC = 67^\circ$, then what is $\angle HCA$?
+591 It's an accute triangle so if we extend CH, the base is orthogonal/perpendicular to AB, so we can find ACH easily: 180-(90+43)=$\boxed{47^\circ}$
