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 $\angle HCA$?
Question:
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 $\angle HCA$?
Answer:
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47^\circ
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 $\angle HCA$?
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