Question

In the figure, $O$ is the centre of the circle. If $\angle ADC={140}^o$, then what is the value of $x$?

• A. ${35}^o$
• B. ${55}^o$
• C. ${60}^o$
• D. ${50}^o$

Answer 1

The correct option is D ${50}^o$

$OA=OC\Rightarrow \angle OAC=\angle ACO=X$

$\angle ADC+\angle AMC={180}^{\circ }$ (cyclic quadrilateral )

$\angle AMC={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

$\angle ADC=2\angle AMC=2\times {40}^{\circ }={80}^{\circ }$

In $△AOC$

$\angle ADC+\angle OAC+\angle OCA={180}^{\circ }$

$x+{80}^{\circ }+x={180}^{\circ }$

$2x={100}^{\circ }$

$x={50}^{\circ }$