Question

Question 14
A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment.

Answer 1

Given, AB is a chord of a circle, which is equal to the radius of the circle, i.e., AB = BO
Join OA, AC and BC.
Since OA = OB = radius of circle i.e.,AB = OA=OB
Thus, $\Delta OAB$ is an equilateral triangle.
$\Rightarrow \angle AOB={60}^{\circ }$ [each angle of an equilateral triangle is ${60}^{\circ }$]

By using the theorem, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$i.e.,\angle AOB=2\angle ACB$
$\angle ACB=\frac{{60}^{\circ }}{2}={30}^{\circ }$
The angle subtended by the chord AB at a point in the major segment is ${30}^{\circ }$ .