Question

$\text{In the given figure,O is the centre of the circle. If}\angle \text{ACB}={60}^{\circ }\text{then find}\angle OAB.$

• A. ${120}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is D

${30}^{\circ }$

$Given\angle AOB={120}^{\circ }$
(The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)
OA = OB (Radius of the circle)



$x+x+{120}^{\circ }={180}^{\circ }$
$\Rightarrow 2x={60}^{\circ }$
$\Rightarrow x={30}^{\circ }$