Question

$ABCD$ is a cyclic quadrilateral. Then, find $\angle x^{\circ }$ as given in the figure.

• A. ${50}^{\circ }$
• B. ${80}^{\circ }$
• C. ${90}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

The correct option is D ${100}^{\circ }$

$\angle EDC+\angle ADC={180}^o...\left(Linearpairofangles\right)\angle EDC={80}^o...\left(Given\right)Then,\angle EDC+\angle ADC={180}^o\angle ADC={180}^o-\angle EDC={180}^o-{80}^o={100}^o.ABCDisacyclicquadrilateral.Hence,\angle ADC+\angle ABC={180}^o...\left(Sumofoppositeanglesis{180}^o\right)\Rightarrow \angle ABC={180}^o-\angle ADC={180}^o-{100}^o={80}^o.Now,\angle ABF+\angle ABC={180}^o\angle ABF={180}^o-\angle ABC\angle ABF=\angle x...\left(Given\right)\Rightarrow \angle x={180}^o-{80}^o\Rightarrow \angle x={100}^o.Hence,optionDiscorrect.$