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 = ∠AOB
⇒ ∠ACB = 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°.