As shown in above figure AB and CD are two parallel lines and a line XY passes through them and intersects AB at P & CD at Q. If $∠ D Q X = x^{\circ}$ then find $∠ A P Y$ .

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Solution

Given $∠ D Q X = x^{\circ}$

$∵ ∠ B P Y$ and $∠ D Q X$ are supplementary angles so $∠ B P Y = 180^{\circ} - x^{\circ}$

$∠ B P Y$ and $∠ A P Y$ are supplementary angles.

$\Rightarrow ∠ B P Y + ∠ A P Y = 180^{\circ}$

$\Rightarrow 180^{\circ} - x^{\circ} + ∠ A P Y = 180^{\circ} \Rightarrow ∠ A P Y = x^{\circ}$

Also, $∠ A P Y = ∠ D Q X = x^{\circ}$ (Since they are alternate interior angles)

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