Question

A chord AB of a circle is equal to the radius of the circle. Find the angles subtended by the chord at points on the major arc and the minor arc.

Answer 1

In $\Delta OAB$,
AB = OA = OB = radius of the circle.
$\therefore \Delta OAB$ is an equilateral triangle.
Therefore, each interior angle of this triangle will be equal to ${60}^{\circ }$.
$\therefore \angle AOB={60}^{\circ }$.

Since 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, we have
$\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }.$

Now in the cyclic quadrilateral ACBD,
$\angle ACB+\angle ADB={180}^{\circ }.$ [Opposite angles in a cyclic quadrilateral are supplementary]
$\therefore \angle ADB={180}^{\circ }-\angle ACB={180}^{\circ }-{30}^{\circ }={150}^{\circ }$

Therefore, the angles subtended by the chord AB at a point on the major arc and the minor arc are ${30}^{\circ }$ and ${150}^{\circ }$ respectively.