Question

$ABCD$ is a rectangle if $\angle BPC={124}^{\circ }$. Calculate$\angle ADP$

• A. ${62}^{\circ }$
• B. ${56}^{\circ }$
• C. ${28}^{\circ }$
• D. None of these

Answer 1

Answer: (C)

${28}^{\circ }$

Given, $\angle BPC={124}^o$

From the diagram, we get

$AC=BD$ and $AP=PC=BP=PD$ ....Diagonals of a\quad rectangle are equal and bisect eachother

$\angle APD=\angle BPC={124}^o$ .... Vertically opposite angles

Here $△APD$ is a isosceles triangle, since $AP=PD$

So in $△APD,\angle PAD=\angle PDA$ ....Base angles are equal in isosceles triangle

In $△APD$, we have

$\angle PAD+\angle ADP+\angle DPA={180}^o$

$\Rightarrow 2\angle ADP+\angle DPA={180}^o$

$\Rightarrow 2\angle ADP+{124}^o={180}^o$

$\Rightarrow 2\angle ADP={180}^o-{124}^o$

$\Rightarrow 2\angle ADP={56}^o$

$\Rightarrow \angle ADP=\frac{56}{2}$

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