Question

In the given figure, $O$ is the centre of the circle. If $\angle AOB={140}^0$ and $\angle OAC={50}^0$; find:
$\angle CBA$

Answer 1

Here $\angle ACB=\frac{1}{2}$Reflex$\left(\angle AOB\right)$

$=\frac{1}{2}({360}^{\circ }-{140}^{\circ })=\frac{220}{2}={110}^{\circ }$

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

Now, $OA=OB$ (Radii of same circle)

$\therefore \angle OBA=\angle OAB=\frac{{180}^{\circ }-{140}^{\circ }}{2}={20}^{\circ }$

$\therefore \angle CAB={50}^{\circ }-{20}^{\circ }={30}^{\circ }$

In $△CAB,$

$\angle CBA-{180}^{\circ }-{110}^{\circ }-{30}^{\circ }={40}^{\circ }$

$\therefore \angle OBC=\angle CBA+\angle OBA={40}^{\circ }+{20}^{\circ }={60}^{\circ }$