$A$ 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.
(A) Let the chord be $AB$ and the center of the circle be $O$. Since the chord $AB$ is equal to the radius of the circle,we have $OA = OB = AB$.
Therefore,$\Delta AOB$ is an equilateral triangle.
Since each angle of an equilateral triangle is $60^{\circ}$,we have $\angle AOB = 60^{\circ}$.
The reflex angle subtended by the chord at the center is $\text{reflex } \angle AOB = 360^{\circ} - 60^{\circ} = 300^{\circ}$.
The angle subtended by the chord at a point $C$ on the minor arc is half of the reflex angle subtended at the center: $\angle ACB = \frac{1}{2} \times 300^{\circ} = 150^{\circ}$.
The angle subtended by the chord at a point $D$ on the major arc is half of the angle subtended at the center: $\angle ADB = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.
Thus,the angle subtended by the chord on the minor arc is $150^{\circ}$ and on the major arc is $30^{\circ}$.
Therefore,$\Delta AOB$ is an equilateral triangle.
Since each angle of an equilateral triangle is $60^{\circ}$,we have $\angle AOB = 60^{\circ}$.
The reflex angle subtended by the chord at the center is $\text{reflex } \angle AOB = 360^{\circ} - 60^{\circ} = 300^{\circ}$.
The angle subtended by the chord at a point $C$ on the minor arc is half of the reflex angle subtended at the center: $\angle ACB = \frac{1}{2} \times 300^{\circ} = 150^{\circ}$.
The angle subtended by the chord at a point $D$ on the major arc is half of the angle subtended at the center: $\angle ADB = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.
Thus,the angle subtended by the chord on the minor arc is $150^{\circ}$ and on the major arc is $30^{\circ}$.
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