question_answer

In a quadrilateral ABCD, the bisectors of \[\angle \mathbf{A}\] and \[\angle B\] meet at O. If \[\angle C={{80}^{{}^\circ }}\]and \[\angle \mathbf{D}=\mathbf{12}{{\mathbf{0}}^{{}^\circ }}\], then measure of \[\angle \mathbf{AOB}\] is

A)  \[{{40}^{{}^\circ }}\]                                   

B)  \[{{60}^{{}^\circ }}\]

C)  \[{{80}^{{}^\circ }}\]                          

D)         \[{{100}^{{}^\circ }}\]

Correct Answer: D

Solution :

(d): \[\angle A+\angle B+\angle C+\angle D={{360}^{{}^\circ }}\] \[\Rightarrow \]\[\angle A+\angle B+\angle {{80}^{{}^\circ }}+\angle {{120}^{{}^\circ }}={{360}^{{}^\circ }}\] \[\Rightarrow \]\[\angle A+\angle B={{360}^{{}^\circ }}-{{80}^{{}^\circ }}-{{120}^{{}^\circ }}={{160}^{{}^\circ }}\] In \[\Delta AOB,\angle OAB+\angle OBA+\angle AOB={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\frac{\angle A}{2}+\frac{\angle B}{2}+\angle AOB={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\frac{1}{2}\left( \angle A+\angle B \right)+\angle AOB={{180}^{{}^\circ }}\] \[\Rightarrow \] \[\frac{1}{2}\times {{160}^{{}^\circ }}+\angle AOB={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\angle AOB={{180}^{{}^\circ }}-{{80}^{{}^\circ }}={{100}^{{}^\circ }}\]