jmc

This question has a sincere need for a diagram!



Since we are considering the incenter, $△BYX$ is isosceles, and indeed is equilateral. Therefore $\angle BYX=60^{\circ }$. This tells us $$180^{\circ }=\angle AYB=\angle AYX+\angle BYX=\angle AYX+60^{\circ }.$$Solving gives $\angle AYX=\boxed{120^{\circ }}$.