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.

Let BC be chord, which is equal to the radius. Join OB and OC.



Given:
BC = OB = OC
\(\therefore, \) \(\triangle{OBC}\) is an equilateral traingle.
\(\therefore \) \(\angle{BOC}\) = \(60^\circ\)
But, we know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any part of the circle.

\(\therefore \) \(\angle{BAC}\) = (\(\frac{1}{2} \) ) \(\angle{BOC}\)
\(\Rightarrow \) \(\angle{BAC}\) = (\(\frac{1}{2} \) ) \(60^\circ\) = \(30^\circ\)

Also, here, ABMC is a cyclic quadrilateral.

\(\therefore \) \(\angle{BAC}\) + \(\angle{BMC}\) = \(180^\circ\)
(Since, in a cyclic quadrilateral the sum of opposite angles is \(180^\circ\)).
\(\therefore \) \(\angle{BMC}\) = \(180^\circ\) - \(30^\circ\) = \(150^\circ\)

Hence, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is \(150^\circ\).