A chord of a circle is equal to its radius. F | vedclass

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(N/A) Let the chord be $AB$ and the center of the circle be $O$. Since the chord is equal to the radius,we have $AB = OA = OB$.Therefore,$	riangle OA

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(N/A) Let the chord be $AB$ and the center of the circle be $O$. Since the chord is equal to the radius,we have $AB = OA = OB$.
Therefore,$\triangle OAB$ is an equilateral triangle.
Since each angle of an equilateral triangle is $60^{\circ}$,we have $\angle AOB = 60^{\circ}$.
According to the circle theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Thus,$\angle AOB = 2 \angle ACB$,where $C$ is a point on the major arc.
Therefore,$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.

$A$ chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in the major segment.

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