Question

In the figure, $\angle$ABC is a circle with centre O. If $\angle$AOB$={82}^o$, find x.

Answer 1

Given : $\angle AOB={82}^o$

$\angle ACB=x$

$AOB$ is an angle subtended by the arc $AB$ at the centre of the circle.

$\angle ACB$ is an angle subtended by the same arc $AB$ on the circle.

By applying the theorem,

"Angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle".

$⟹\angle AOB=2\angle ACB$

$⟹\angle ACB=\frac{1}{2}\times \angle AOB$

$⟹\angle ACB=\frac{1}{2}\times {82}^o={41}^o$

$\therefore \angle ACB={41}^o$

Hence, $x={41}^o$