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

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(B) To find $\angle BPC$,draw a line $EF$ passing through point $P$ such that $EF \parallel AB$. Since $AB \parallel CD$ and $EF \parallel AB$,it foll

Vedclass · Accepted Answer

$90$

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(B) To find $\angle BPC$,draw a line $EF$ passing through point $P$ such that $EF \parallel AB$. Since $AB \parallel CD$ and $EF \parallel AB$,it follows that $EF \parallel CD$.$1$. Consider the parallel lines $AB$ and $EF$. Since $BP$ is a transversal,the interior angles on the same side sum to $180^{\circ}$.$\angle PBA + \angle BPE = 180^{\circ}$$140^{\circ} + \angle BPE = 180^{\circ}$$\angle BPE = 180^{\circ} - 140^{\circ} = 40^{\circ}$.$2$. Consider the parallel lines $CD$ and $EF$. Since $CP$ is a transversal,the interior angles on the same side sum to $180^{\circ}$.$\angle PCD + \angle CPE = 180^{\circ}$$130^{\circ} + \angle CPE = 180^{\circ}$$\angle CPE = 180^{\circ} - 130^{\circ} = 50^{\circ}$.$3$. Now,$\angle BPC = \angle BPE + \angle CPE = 40^{\circ} + 50^{\circ} = 90^{\circ}$.

In the given figure,if $AB \parallel CD$,$\angle PCD = 130^{\circ}$ and $\angle PBA = 140^{\circ}$,then find $\angle BPC$. (in $^{\circ}$)

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