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(A) Since ray $YW$ stands on line $XYZ$,$\angle WYZ + \angle WYX = 180^{\circ}$ (Linear pair of angles).Given that $\angle WYZ : \angle WYX = 1 : 2$,l

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$90$

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(A) Since ray $YW$ stands on line $XYZ$,$\angle WYZ + \angle WYX = 180^{\circ}$ (Linear pair of angles).Given that $\angle WYZ : \angle WYX = 1 : 2$,let $\angle WYZ = x$ and $\angle WYX = 2x$.Therefore,$x + 2x = 180^{\circ} \implies 3x = 180^{\circ} \implies x = 60^{\circ}$.So,$\angle WYZ = 60^{\circ}$ and $\angle WYX = 120^{\circ}$.Ray $YQ$ is the angle bisector of $\angle WYZ$,so $\angle WYQ = \frac{1}{2} 	imes \angle WYZ = \frac{1}{2} 	imes 60^{\circ} = 30^{\circ}$.Ray $YP$ is the angle bisector of $\angle WYX$,so $\angle WYP = \frac{1}{2} 	imes \angle WYX = \frac{1}{2} 	imes 120^{\circ} = 60^{\circ}$.Finally,$\angle PYQ = \angle WYQ + \angle WYP = 30^{\circ} + 60^{\circ} = 90^{\circ}$.

In the following figure,ray $YW$ stands on a line $XYZ$. $\angle WYZ : \angle WYX = 1 : 2$. Ray $YQ$ and ray $YP$ are angle bisectors of $\angle WYZ$ and $\angle WYX$ respectively. Find $\angle PYQ$. (in $^{\circ}$)

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