Question

In the given figure, AB is the chord to the circle having centre at O. If AC = CB and $\angle AOC={50}^{\circ }$, then $\angle CAO$ is equal to.

• A. ${30}^{\circ }$
• B. ${50}^{\circ }$
• C. ${40}^{\circ }$

Answer 1

The correct option is C ${40}^{\circ }$

In the given figure AB is the chord and AC = CB.
Since OC is bisecting the chord AB, OC will be perpendicular to AB.
(The line drawn through the center of a circle to bisect a chord is perpendicular to the chord)
Hence, $\angle OCA={90}^{\circ }$

Now in $\Delta OAC$
$\angle OCA={90}^{\circ }$
$\angle AOC={50}^{\circ }$
So, $\angle CAO={180}^{\circ }-({50}^{\circ }+{90}^{\circ })={40}^{\circ }$