Question

ABCD is a rectangle, if $\angle BPC={124}^{\circ }$

Calculate $\angle BAP$

• A. ${56}^{\circ }$
• B. ${72}^{\circ }$
• C. ${45}^{\circ }$
• D. ${62}^{\circ }$

Answer 1

The correct option is D

${62}^{\circ }$

Diagonals of a rectangle are equal and bisect each other.

Let $\angle PBC=\angle PCB=x$

$\Rightarrow \angle PBC+\angle BPC+\angle PCB={180}^{\circ }$

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

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

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

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

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

$\Rightarrow \angle PBC=\angle ADP\left[Alternateangles\right]$

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

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

$\Rightarrow PA=PB$

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

$\Rightarrow \angle BAP=\frac{1}{2}\times ({180}^{\circ }-{56}^{\circ })$

$\Rightarrow \angle BAP=\frac{1}{2}\times {124}^{\circ }={62}^{\circ }$