Question

Question 2
A chord AB of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the major arc and also at a point on the minor arc.

Answer 1

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

We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
$\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\left({60}^{\circ }\right)={30}^{\circ }$

In cyclic quadrilateral ACBD,
$\angle ACB+\angle ADB={180}^{\circ }$ (Opposite angles in a cyclic quadrilateral are supplementary)
$\therefore \angle ADB={180}^{\circ }-{30}^{\circ }={150}^{\circ }$
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are ${30}^{\circ }$ and ${150}^{\circ }$ respectively