Question

In the given figure $m\angle A={84}^{\circ },m\angle B={80}^{\circ },$ then

• A. $m\angle C={96}^{\circ },m\angle D={100}^{\circ }$
• B. $m\angle C={100}^{\circ },m\angle D={96}^{\circ }$
• C. $m\angle C={100}^{\circ },m\angle D={100}^{\circ }$
• D. $m\angle C={96}^{\circ },m\angle D={96}^{\circ }$

Answer 1

The correct option is A $m\angle C={96}^{\circ },m\angle D={100}^{\circ }$


Here we have two quadrilaterals. $APQB$ inscribed in the circle on the left and $PCDQ$ inscribed in the circle on the right.
We know that, opposite angles of an inscribed quadrilateral are supplementary, so using this theorem on quadrilateral $APQB.$
$\therefore m\angle B+m\angle APQ={180}^{\circ }$
$\Rightarrow m\angle APQ={180}^{\circ }-{80}^{\circ }={100}^{\circ }$
$\because m\angle APQ+m\angle CPQ={180}^{\circ }$
$\Rightarrow m\angle CPQ={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

Similarly, $\angle BQP={180}^{\circ }-{84}^{\circ }={96}^{\circ }$
and $\angle DQP={180}^{\circ }-{96}^{\circ }={84}^{\circ }$

Now using the same theorem on quadrilateral $PCDQ.$
$\therefore m\angle QDC+m\angle P={180}^{\circ }$
$\Rightarrow m\angle QDC={180}^{\circ }-{80}^{\circ }={100}^{\circ }$
and $m\angle PCD+m\angle PQD={180}^{\circ }$
$m\angle PCD={180}^{\circ }-{84}^{\circ }={96}^{\circ }$

$\therefore \angle C={96}^{\circ }$&$\angle D={100}^{\circ }$