Question

In the figure, A, B, C are points on a circle with centre O. If $\angle$ACB$={40}^o$, find $\hat{x}=\angle AOB$.

Answer 1

$A,B,C$ are the points on a circle with centre $O$.

Join $AB,BC$ and $AC$ which forms triangle $ABC$.

$\angle ACB={40}^o$ ....... $\left(i\right)$ [Given]

Let $\angle AOB=\hat{x}$

$\angle AOB$ is an angle made by the arc $AB$ at the centre and $\angle ACB$ is another angle made by the same arc $AB$ at point $C$ on the circle.

Now, we use the theorem "Angle made by an arc at the centre is double of the angle made by the same arc on the circle".

We have $\hat{x}=2\hat{C}$
$\therefore \hat{x}=2\left({40}^o\right)={80}^o$.