jmc

First, we build a diagram:



Since altitudes $\overline{AX}$ and $\overline{BY}$ intersect at $H$, point $H$ is the orthocenter of $△ABC$. Therefore, the line through $C$ and $H$ is perpendicular to side $\overline{AB}$, as shown. Therefore, we have $\angle HCA=\angle ZCA=90^{\circ }-43^{\circ }=\boxed{47^{\circ }}$.