Question

In quadrilateral ABCD, AP and BP are bisectors of $\angle A$ and $\angle B$ respectively, then the value of $x$ is :

• A. ${60}^{\circ }$
• B. ${85}^{\circ }$
• C. ${95}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

Answer: (C)

${95}^{\circ }$

$AP$ and $BP$ are bisectors of two adjacent angles A and B of a quadrilateral $ABCD.$

We know that the sum of all the angles of a quadrilateral is $360°.$

$\Rightarrow \angle A+\angle B+\angle C+\angle D=360°,$ $\angle A+\angle B=360°-(\angle D+\angle C)$

$\therefore$ In $△PAB;$

$\Rightarrow \angle APB+\angle PAB+\angle PBA=180°$ [ Angle sum property ]

$\Rightarrow \angle APB+\frac{1}{2}\angle A+\frac{1}{2}\angle B=180°$

$\therefore AP$ and $BP$ are the bisectors of two adjacent angles A and B.

$\therefore 2\angle APB+\angle A+\angle B=360°$

$\Rightarrow 2\angle APB=360°-(\angle A+\angle B)$

$\therefore 2\angle APB=\angle C+\angle D$

$\Rightarrow \angle APB=\frac{{130}^o+{60}^o}{2}=95°$

$\therefore x=95°.$

Hence, the answer is $95°.$