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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.
Answer :

Let BC be chord, which is equal to the radius. Join OB and OC.
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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 \(150^\circ\).