Problem 487

Let $\angle XYW=\theta$. Since $\triangle XYW$ is isosceles with $WX=WY$, then $\angle YXW=\angle XYW=\theta$. Since the sum of the angles in $\triangle XYW$ is $180^{\circ}$, then $\angle XWY=180^{\circ}-2\theta$. Since $\angle XWY+\angle ZWY=180^{\circ}$, then $\angle ZWY=180^{\circ}-(180^{\circ}-2\theta)=2\theta$. Since $\triangle YWZ$ is isosceles with $YW=YZ$, then $\angle YZW=\angle ZWY=2\theta$. Since $\triangle XYZ$ is isosceles, with $XY=XZ$, then $\angle XYZ=\angle XZY=2\theta$. Since the sum of the angles in $\triangle XYZ$ is $180^{\circ}$, then $\angle XYZ+\angle XZY+\angle YXZ=180^{\circ}$ or $2\theta+2\theta+\theta=180^{\circ}$, or $5\theta=180^{\circ}$, or $\theta=36^{\circ}$.