Question

In the given figure, if $\angle \text{OAB}={40}^{\circ }$, then $\angle \text{ACB}$ = ___.

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

Answer 1

The correct option is A

${50}^{\circ }$

Note that OA = OB=radius of the circle.
Since angles opposite to equal sides are equal, $\angle OAB=\angle OBA={40}^{\circ }.$
Using angle sum property in $△OAB,$ we have
$\angle AOB={180}^{\circ }-({40}^{\circ }+{40}^{\circ })={100}^{\circ }.$
Since the angle subtended by an arc at the centre is double the angle subtended by it at any remaining part of the circle, we have
$\angle ACB=\frac{\angle AOB}{2}={50}^{\circ }.$