jmc

First, we build a diagram:



From quadrilateral $CXHY$, we have $$\begin{array}{rl}\angle ACB=\angle XCY& =360^{\circ }-\angle HXC-\angle YHX-\angle CYH\\ & =360^{\circ }-90^{\circ }-\angle YHX-90^{\circ }\\ & =180^{\circ }-\angle YHX.\end{array}$$Since $\angle YHX=\angle AHB$, we have $\angle ACB=180^{\circ }-\angle YHX=\boxed{48^{\circ }}$.