Question

In the given figure, $\angle A={60}^{\circ }$ and $\angle ABC={80}^{\circ }$, then $\angle DPC$ and $\angle BQC$ are respectively ___.

• A. ${60}^{\circ },{30}^{\circ }$
• B. ${40}^{\circ },{40}^{\circ }$
• C. ${40}^{\circ },{20}^{\circ }$
• D. ${20}^{\circ },{60}^{\circ }$

Answer 1

The correct option is C

${40}^{\circ },{20}^{\circ }$

In a cyclic quadrilateral, exterior angle is equal to opposite interior angle.
So, in cyclic quadrilateral ABCD, we have,

$\angle PDC=\angle ABC$ and $\angle DCP=\angle A$

$\angle PDC={80}^{\circ }$ and $\angle DCP={60}^{\circ }$ [Given, $\angle ABC={80}^{\circ }$ and $\angle A={60}^{\circ }$ ]

In $\Delta DPC$, we have

$\angle DPC={180}^{\circ }-(\angle PDC+\angle DCP)$

$\Rightarrow \angle DPC={180}^{\circ }-({80}^{\circ }+{60}^{\circ })={40}^{\circ }$

Similarly, we have
$\angle QBC=\angle ADC$ and $\angle BCQ=\angle BAD$

[$\angle ADC+\angle ABC={180}^{\circ }$ (Opposite angle sum of cyclic quadrilateral), $\angle ABC={80}^{\circ }$ and $\angle A={60}^{\circ }$]

$\angle QBC={180}^{\circ }-\angle ABC$

$\angle QBC={180}^{\circ }-{80}^{\circ }={100}^{\circ }and\angle BCQ=\angle BAD={60}^{\circ }$

Now, in $\Delta BQC$, we have

$\angle BQC={180}^{\circ }-(\angle QBC+\angle BCQ)$

$\Rightarrow \angle BQC={180}^{\circ }-({100}^{\circ }+{60}^{\circ })={20}^{\circ }$