In the given figure,if AB parallel CD,angle A | vedclass

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(C) Given $AB \parallel CD$. Since $AB \parallel CD$ and $XY$ is a transversal,the alternate interior angles are equal. Therefore,$\angle C Y X = \ang

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$x=80^{\circ}, y=60^{\circ}$

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(C) Given $AB \parallel CD$. Since $AB \parallel CD$ and $XY$ is a transversal,the alternate interior angles are equal. Therefore,$\angle C Y X = \angle AXY = 80^{\circ}$. Since $CYD$ is a straight line,$\angle C Y X + \angle X Y Z = 180^{\circ}$ (linear pair). $80^{\circ} + x = 180^{\circ} \implies x = 100^{\circ}$. Wait,looking at the figure,$\angle X Y Z$ is an interior angle. Since $AB \parallel CD$,$\angle AXY + \angle X Y C = 180^{\circ}$ (consecutive interior angles). $80^{\circ} + \angle X Y C = 180^{\circ} \implies \angle X Y C = 100^{\circ}$. Since $CYD$ is a straight line,$\angle X Y C + \angle X Y Z = 180^{\circ} \implies 100^{\circ} + x = 180^{\circ} \implies x = 80^{\circ}$. Now,for $\angle XZD = 140^{\circ}$,since $CYD$ is a straight line,$\angle X Z Y + \angle X Z D = 180^{\circ} \implies \angle X Z Y + 140^{\circ} = 180^{\circ} \implies \angle X Z Y = 40^{\circ}$. In $	riangle XYZ$,the sum of angles is $180^{\circ}$. $\angle X Y Z + \angle X Z Y + \angle Y X Z = 180^{\circ} \implies 80^{\circ} + 40^{\circ} + y = 180^{\circ} \implies 120^{\circ} + y = 180^{\circ} \implies y = 60^{\circ}$. Thus,$x = 80^{\circ}$ and $y = 60^{\circ}$.

In the given figure,if $AB \parallel CD$,$\angle AXY = 80^{\circ}$ and $\angle XZD = 140^{\circ}$,then find $x$ and $y$.

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