jmc

Since $△ABC$ and $△PQR$ are equilateral, then $\angle ABC=\angle ACB=\angle RPQ=60^{\circ }$.

Therefore, $\angle YBP=180^{\circ }-65^{\circ }-60^{\circ }=55^{\circ }$ and $\angle YPB=180^{\circ }-75^{\circ }-60^{\circ }=45^{\circ }$.

In $△BYP$, we have $\angle BYP=180^{\circ }-\angle YBP-\angle YPB=180^{\circ }-55^{\circ }-45^{\circ }=80^{\circ }$.

Since $\angle XYC=\angle BYP$, then $\angle XYC=80^{\circ }$.

In $△CXY$, we have $\angle CXY=180^{\circ }-60^{\circ }-80^{\circ }=40^{\circ }$.

So our final answer is $\boxed{40}$ degrees.