Question

In the given figure, O is the centre of the circle . $\angle OAB$ and $\angle OCB$ are ${30}^{\circ }$ and ${40}^{\circ }$ respectively . Find $\angle AOC$. Show your steps of working

Answer 1

Let $\angle OAC=\angle OCA=x$

$\therefore \angle AOC={180}^{\circ }-2x$

Also,$\angle BAC={30}^{\circ }+x$

and $\angle BCA={40}^{\circ }+x$

In $△ABC$,

$\angle ABC={180}^{\circ }-\angle BAC-\angle BCA$

$={180}^{\circ }-({30}^{\circ }+x)-({40}^{\circ }+x)={110}^{\circ }-2x$

Now $\angle AOC=2\angle ABC$

(Angle at the centre is double the angle at the circumference subtended by the same chord)

$\Rightarrow {180}^{\circ }-2x=2({110}^{\circ }-2x)$

$\Rightarrow 2x={40}^{\circ }$

$\therefore x={20}^{\circ }$

$\therefore \angle AOC={180}^{\circ }-2\times {20}^{\circ }={180}^{\circ }-{40}^{\circ }={140}^{\circ }$