2. 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.

So, \(\angle{BOC}\) = \(60^\circ\)

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.

So, \(\angle{BAC}\) = (1/2) \(\angle{BOC}\)

i.e., \(\angle{BAC}\) = (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 \(180^\circ\)).

So, \(\angle{BMC}\) = \(180^\circ\) - \(30^\circ\) = \(150^\circ\)

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