Question

In a quadrilateral ABCD, bisectors of A and B intersect at O such that $\angle AOB={75}^{\circ }$, then write the value $\angle C+\angle D$.

Answer 1

In quadrilateral ABCD,

Bisectors of $\angle Aand\angle B$ meet at O and $\angle AOB={75}^{\circ }$

In AOB , $\angle AOB={75}^{\circ }$

$\therefore \angle OAB+\angle OBA={180}^{\circ }-{75}^{\circ }={105}^{\circ }$

But OA and OB are the bisectors of $\angle Aand\angle B.$

$\therefore \angle A+\angle B=2\times {105}^{\circ }={210}^{\circ }$

But $\angle A+\angle B+\angle C+\angle D={360}^{\circ }$ (Sum of angles of a quad.)

$\therefore {210}^{\circ }+\angle C+\angle D={360}^{\circ }$

$\Rightarrow \angle C+\angle D={360}^{\circ }-{210}^{\circ }={150}^{\circ }$

Hence $\angle C+\angle D={150}^{\circ }$