Question

In the figure, O is the centre of the circle. If $\angle APB={50}^o$, find $\angle AOBand\angle OAB$.

Answer 1

Arc AB, subtends $\angle AOB$ at the centre and $\angle APB$ at the remaining part of the circle

$\therefore \angle AOB=2\angle APB=2\times {50}^o={100}^o$

Join AB

$\Delta AOB$ is an isosceles triangle in which

$OA=OB$

$\therefore \angle OAB=\angle OBA$

But $\angle AOB={100}^o$

(Since angle subtended by an arc of a circle at its centre is twice the angle subtended by it at any point of the alternate segment of circle)

$\therefore \angle OAB+\angle OBA={180}^o-{100}^o={80}^o$

$\Rightarrow 2\angle OAB={80}^o$ as $\angle OAB=\angle OBA$

$\therefore \angle OAB=\frac{{80}^o}{2}={40}^o$