question_answer

In a quadrilateral ABCD, If AO and BO are the bisectors of \[\angle \mathbf{A}\] and \[\angle B\] respectively \[\angle \mathbf{C}=\mathbf{3}{{\mathbf{0}}^{{}^\circ }}\]and\[\angle \mathbf{D}=\mathbf{7}{{\mathbf{0}}^{{}^\circ }}\]. Then, \[\angle \mathbf{AOB}=\]?

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

B)  \[{{50}^{{}^\circ }}\]

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

D)  \[{{100}^{{}^\circ }}\]

Correct Answer: B

Solution :

(b): Sum of the angles of quadrilateral is\[{{360}^{{}^\circ }}\], \[\therefore \]\[\angle A+\angle B+{{30}^{{}^\circ }}+{{70}^{{}^\circ }}={{360}^{{}^\circ }}\] \[\therefore \]\[\angle A+\angle B={{260}^{{}^\circ }}\] \[\Rightarrow \]\[\frac{1}{2}\angle A+\frac{1}{2}\angle B={{130}^{{}^\circ }}\] \[\therefore \angle AOB=(180{}^\circ -130{}^\circ )=50{}^\circ \]