Question

In the given figure, AB is the chord to the circle having centre at O. If AC = CB and $\angle AOC={60}^{\circ }$, then $\angle CAO$ is equal to__.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$

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 $\Delta OAC$
$\angle OCA={90}^{\circ }$
$\angle AOC={60}^{\circ }$
So, $\angle CAO={180}^{\circ }-({60}^{\circ }+{90}^{\circ })={30}^{\circ }$