Question

In the given figure if 'O' is the centre of the circle then $\angle AOB$ =___

Answer 1

Give $\angle OAC={20}^{\circ }$ and $\angle OBC={30}^{\circ }$

Let us draw aa line joining 'O' and 'C'

Now in $\Delta OAC$

OA = OC

$\therefore \angle OCA=\angle OAC$

$\therefore \angle OCA={20}^{\circ }$

Similarly in $\Delta BOC$

OB = OC

$\therefore \angle OCB=\angle OBC$

$\Rightarrow \angle OCB={30}^{\circ }$

$\therefore \angle ACB=\angle OCA+\angle OCB$

$={20}^{\circ }+{30}^{\circ }$

$={50}^{\circ }$

Now $\angle AOB=2\times \angle ACB$

($because$ The angle subtends by arc at centre is twice that it subtends at any point on the remaining part of circumference)

$\therefore \angle AOB=2\times {50}^{\circ }$

$={100}^{\circ }$