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.

 

Here OP = OQ = PQ

Hence ∆PQO becomes an equilateral triangle.

Draw 2 points C and D on the circle such that they lie on the major arc and minor arc, respectively.

Since ∆PQO is an equilateral triangle, we get ∠POQ = 60°.

For the arc PQ, ∠POQ = 2∠PCQ as we know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

∠PCQ = 1/2 ∠POQ = 1/2 × 60 = 30°

As you can notice the points P, Q, C, and D lie on the circle. Hence PQCD is a cyclic quadrilateral.

We know that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

Therefore,

∠PCQ + ∠PDQ = 180°

30° + ∠PDQ = 180°

∠PDQ = 150°

So, when the chord of a circle is equal to the radius of the circle, the angle subtended by the chord at a point on the minor arc is 150° and also at a point on the major arc is 30°.