Question

$ABCD$ is a quadrilateral. $AO$ and $BO$ are the angle bisectors of angles $A$ and $B$ which meet at $O$. If $\angle C={70}^o,\angle D={50}^o$, find $\angle AOB$.

Answer 1

We have,

ABCD is a quadrilateral.

Given that,

$\angle C={70}^o,\angle D={50}^o$

So,

$\angle A+\angle B+\angle C+\angle D={360}^o$

$(A+B)={360}^o-(C+D)$

$\Rightarrow (A+B)={360}^o-({70}^o+{50}^o)$

$\Rightarrow (A+B)={360}^o-{120}^o$

$\Rightarrow (A+B)={240}^o$

According to given question,

$\angle OAB=\frac{\angle A}{2}$

$\angle OAB=\frac{\angle B}{2}$

Now,

$\angle OAB+\angle OBA=\frac{\angle A+\angle B}{2}={120}^o$

$\Rightarrow \angle AOB+{120}^o={180}^o$

$\Rightarrow \angle AOB={180}^o-{120}^o$

$\Rightarrow \angle AOB={60}^o$

Hence, this is the answer.