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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 $\angle HCA$?

 Feb 28, 2021
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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)=$\boxed{47^\circ}$

 Feb 28, 2021

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