Question

A chord of a circle AB is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc in degree.

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Answer 1

As OA= OB = AB [given]

$\therefore$ $\Delta$OAB is equilateral

$\therefore$ $\angle$AOB = 60${}^{\circ }$

$\angle$ACB = $\frac{1}{2}$×​ $\angle$AOB = $\frac{1}{2}$×60${}^{\circ }$ = 30${}^{\circ }$ [The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle]

Since ADBC is a cyclic quadrilateral,

$\therefore$ $\angle$ADB + $\angle$ACB = 180${}^{\circ }$ [Sum of either pair of opposite angles of a cyclic quadrilateral is 180${}^{\circ }$]

$\angle$ADB + 30${}^{\circ }$ = 180${}^{\circ }$

$\angle$ADB = 150${}^{\circ }$