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# 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 $$150^\circ$$.