Question

Given a circle with O as the centre. Find the value of x.

• A. 80°
• B. 150°
• C. 50°
• D. 40°

Answer 1

The correct option is D

40°

$\Delta AOB$ is an isosceles triangle [OA = OB = Radius]
$\Rightarrow \angle OAB=\angle OBA={50}^{\circ }$ [Angles opposite to equal sides are equal]

In $\Delta AOB,$

$\angle OAB+\angle OBA+\angle AOB={180}^{\circ }$ [Sum of angles of a triangle is ${180}^{\circ }$]

$\angle AOB={180}^{\circ }-\angle OAB-\angle OBA$
$\angle AOB={180}^{\circ }-2\times \angle OAB$
$\angle AOB={180}^{\circ }-2\times 50$
Hence, $\angle AOB={80}^{\circ }$

The angle subtended by an arc of the circle at its centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore, $\angle ACB=\frac{1}{2}\angle AOB={40}^{\circ }$.