Question

In given figure, if $\angle OAB$ = ${40}^{\circ },$ then $\angle ACB$ is equal to:

• A. ${50}^{\circ }$
• B. ${40}^{\circ }$
• C. ${60}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

The correct option is A ${50}^{\circ }$

In the circle $OA=OB=$ Radius

So, $\angle OAB=\angle OBA={40}^{\circ }$

Therefore, $\angle AOB={180}^{\circ }-{40}^{\circ }-{40}^{\circ }={100}^{\circ }$ {Angle sum property}

We know in a circle the angle subtended by an arc at the center is twice that of the angle made on the circle.

Therefore, $\angle ACB=\frac{\angle AOB}{2}={50}^{\circ }$