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.

From figure,

There is a circle with center O and chord AB which is equal to radius of the circle.

In ∆ AOB,

⇒ OA = OB = AB

∴ ∆ OAB is an equilateral triangle.

⇒ ∠AOB = 60° (Each angle of an equilateral triangle equals to 90°)

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.

⇒ ∠AOB = 2∠ACB

⇒ ∠ACB = $\dfrac{1}{2}$ ∠AOB

⇒ ∠ACB = $\dfrac{1}{2}$ x 60° = 30°

⇒ ∠ACB = 30°.

Since, ACBD is a cyclic quadrilateral

⇒ ∠C + ∠D = 180° (Opposite angles of cyclic quadrilateral are supplementary)

⇒ 30° + ∠D = 180°

⇒ ∠D = 180° - 30°

⇒ ∠D = 150°.

Hence, angle subtended on minor arc equals to 150° and angle subtended on major arc equals 30°.