A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

We know that the angle subtended by a chord at the centre is twice the same at the circumference.

Here, $AB$ is a chord. $AB=OA=OB$

$\therefore ∆AOB$ is an equilateral triangle.

$\therefore \angle AOB={60}^{\circ }$

So, we get,

$$\begin{gathered} \angle AOB=2\angle APB \\ {60}^0=2\angle APB \\ \therefore \angle APB={30}^0 \end{gathered}$$

By cyclic quadrilateral property,

$\angle APB+\angle ACB={180}^0$

 $\Rightarrow {30}^0+\angle ACB={180}^0$

$\Rightarrow \angle ACB={150}^0$

Hence, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc are ${150}^o, {30}^o$ respectively.