Question

In the figure, $DE\parallel QR$ and $AP$ & $BP$ are bisectors of $\angle EAB$ and $\angle RBA,$ respectively. Find $\angle APB.$

• A. ${135}^0$
• B. ${120}^0$
• C. ${90}^0$
• D. ${100}^0$

Answer 1

The correct option is C ${90}^0$

Given: $DE\parallel QR$, AP bisects $\angle EAB$, BP bisects $\angle RBA$
Since, $DE\parallel QR$
$\angle EAB+\angle RBA={180}^0$
$\frac{1}{2}(\angle EAB+\angle RBA)={90}^0$
$\angle PAB+\angle PBA={90}^0$ (AP bisects $\angle EAB$, BP bisects $\angle RBA$)
Now, In $△PAB$
$\angle PAB+\angle PBA+\angle APB={180}^0$
${90}^0+\angle APB={180}^0$
$\angle APB={90}^{\circ }$