Question

In the given figure, AO and BO are bisecters of $\angle A$ and $\angle B$ respectively. If $\angle C={125}^{\circ }$ and $\angle D={75}^{\circ }$, then x=___

• A. ${80}^{\circ }$
• B. ${90}^{\circ }$
• C. ${120}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

The correct option is D

${100}^{\circ }$

$\angle A+\angle B+75+125={360}^{\circ }$ (Sum of angles of a quadrilateral)

$\angle A+\angle B={360}^{\circ }-{200}^{\circ }$

$={160}^{\circ }$

AO bisects $\angle A$ and BO bisects $\angle B$ (given)

$\therefore \frac{\angle A+\angle B}{2}=\frac{{160}^{\circ }}{2}={80}^{\circ }$

In $\Delta AOB$

$\frac{\angle A}{2}+\frac{\angle B}{2}+x={180}^{\circ }$ (sum of $\angle s$ of a $\Delta )$

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

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