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 $$150^\circ$$.