Question

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

$\text{Calculate:}\left(1\right)\angle BAP\text{(ii)}\angle ADP$

Answer 1

Diagonals of rectangle are equal and bisect each other.
$\angle PBC=\angle PCB=x\text{(say)}$

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

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

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

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

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

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

$\text{But}\angle PBC=\angle ADP\text{[Alternate}\angle S]$

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

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

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

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

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

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