Question

$ABCD$ is a rectangle , if $\angle BPC={124}^{\circ }$ Calculate : $\left(i\right)\angle BAP\left(ii\right)\angle ADP$

Answer 1

Diagonals of the rectangle are equal and bisect each other.

$\therefore \angle PBC=\angle PCB=x\left(say\right)$ [$\because$ PB $=$ PC]

But $\angle BPC+\angle PBC+\angle PCB={180}^{\circ }$ [Angle sum property]

$\Rightarrow {124}^{\circ }+x+x={180}^{\circ }$

$\Rightarrow 2x={180}^{\circ }-{124}^{\circ }$

$\Rightarrow 2x={56}^{\circ }$

$\Rightarrow x={28}^{\circ }$

$\therefore \angle PBC={28}^{\circ }$

But $\angle PBC=\angle ADP$ [Alternate interior angles ]

$\therefore \angle ADP={28}^{\circ }$

Again $\angle APB={180}^{\circ }-{124}^{\circ }={56}^{\circ }$

Also, $PA=PB$

$\therefore \angle BAP=\frac{1}{2}({180}^{\circ }-\angle APB)$

$=\frac{1}{2}\times ({180}^{\circ }-{56}^{\circ })=\frac{1}{2}\times {124}^{\circ }={62}^{\circ }$

Hence $\left(i\right)\angle BAP={62}^{\circ }\left(ii\right)\angle ADP={28}^{\circ }$