Question

In the given figure, $\angle A={70}^{\circ }$ and $\angle ABC={50}^{\circ }$ , find $\angle DPC$ and $\angle BQC$.

• A. 30^o
• B. 40^o
• C. 20^o
• D. 50^o

Answer 1

The correct option is C

20^o

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

$\angle PDC=\angle ABC$
$\text{and}\angle DCP=\angle A$

$\angle PDC={50}^{\circ }\text{and}\angle DCP={70}^{\circ }[\angle ABC={50}^{\circ }\text{and}\angle A={70}^{\circ }]$

$\text{In}\Delta DPC,\text{we have}$

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

$\Rightarrow \angle DPC={180}^{\circ }-({70}^{\circ }+{50}^{\circ })={60}^{\circ }$

$\text{Similarly, we have}$

$\angle QBC=\angle ADC\text{and}\angle BCQ=\angle A$

[$\angle ADC+\angle ABC={180}^{\circ },\angle ABC={50}^{\circ }\text{and}\angle A={70}^{\circ }$]

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

$\angle QBC={180}^{\circ }-{50}^{\circ }={130}^{\circ }\text{and}\angle BCQ={70}^{\circ }$

$\text{Now, in}\Delta BQC,\text{we have}$

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

$\Rightarrow \angle BQC={180}^{\circ }-({100}^{\circ }+{70}^{\circ })={10}^{\circ }$