In the figure,DE parallel QR and AP and BP ar | vedclass

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(B) Given that $DE \parallel QR$ and line $n$ is a transversal intersecting them.Since the sum of consecutive interior angles on the same side of a tr

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$90$

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(B) Given that $DE \parallel QR$ and line $n$ is a transversal intersecting them.Since the sum of consecutive interior angles on the same side of a transversal is $180^{\circ}$,we have:$\angle EAB + \angle RBA = 180^{\circ}$.Since $AP$ is the bisector of $\angle EAB$,$\angle PAB = \frac{1}{2} \angle EAB$.Since $BP$ is the bisector of $\angle RBA$,$\angle PBA = \frac{1}{2} \angle RBA$.Adding these,we get $\angle PAB + \angle PBA = \frac{1}{2} (\angle EAB + \angle RBA) = \frac{1}{2} (180^{\circ}) = 90^{\circ}$.In $\Delta APB$,the sum of angles is $180^{\circ}$:$\angle APB + \angle PAB + \angle PBA = 180^{\circ}$.Substituting the value,$\angle APB + 90^{\circ} = 180^{\circ}$.Therefore,$\angle APB = 180^{\circ} - 90^{\circ} = 90^{\circ}$.

In the figure,$DE \parallel QR$ and $AP$ and $BP$ are bisectors of $\angle EAB$ and $\angle RBA$ respectively. Find $\angle APB.$ (in $^{\circ}$)

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