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 monor arc and also at a point on the major arc.

Explanation

Given, AB is a chord of a circle, which is equal to the radius of the circle, i.e.,
AB=BO …(i)
Join OA, AC and BC.
Since, OA =OB= Radius of circle
`:. OA=AB=BO` [from Eq. (i)]
Thus, ` Delta OAB` is an equilateral triangle.
`rArr angle AOB=60^(@)` [each angle of an equilateral triangle is `60^(@)`]

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., `angleAOB=2 angleACB`
`rArr angleACB=60^(@)/2=30^(@)`