Question : The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is:
Option 1: $75^{\circ}$
Option 2: $60^{\circ}$
Option 3: $150^{\circ}$
Option 4: $120^{\circ}$
Correct Answer: $150^{\circ}$
Solution : Chord AB = Radius In $\triangle$OAB, OA = BO = Radius ⇒ $\triangle$OAB is an equilateral triangle. ⇒ $\angle$AOB = $60^{\circ}$ Since the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. ⇒ $\angle$AOB = 2 × $\angle$ACB ⇒ $\angle$ACB = $\frac{60^{\circ}}{2}$ = $30^{\circ}$ Since ACBD is a cyclic quadrilateral and opposite angles of a cyclic quadrilateral are supplementary, $\angle$ACB + $\angle$ADB = $180^{\circ}$ ⇒ $\angle$ADB = $180^{\circ}- 30^{\circ}=150^{\circ}$ Hence, the correct answer is $150^{\circ}$.
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