Question

In the given figure if $\angle ADC={110}^{\circ },$ then $\angle ABC=$

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${60}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is A ${40}^{\circ }$

We know that, the measure of the angle subtended by an arc at the center is double the measure of the angle subtended by it at any point on the remaining part of the circle.
$\Rightarrow m\angle AOC$ (major arc) $=2\angle ADC$
$\Rightarrow m\angle AOC$ (major arc) $={220}^{\circ }$

and $m\angle AOC$ (major arc)$+m\angle AOC$ (minor arc)$={360}^{\circ }$
$\therefore m\angle AOC$ (minor arc) $={360}^{\circ }-{220}^{\circ }={140}^{\circ }(\because m\angle AOC$ (major arc) $={220}^{\circ })$

Now, $OABC$ is an inscribed quadrilateral and opposite angles of an inscribed quadrilateral are supplementary)
$\therefore m\angle AOC$ (minor arc) $+\angle ABC={180}^{\circ }$
$\Rightarrow \angle ABC={180}^{\circ }-{140}^{\circ }={40}^{\circ }$