### Solution:

A chord of a circle is equal to the radius of the circle. The angle subtended by the chords at a point on the minor arc and also at a point on the major arc is 30o.It is given that chord of a circle is equal to its radius.

Let O be the centre of the circle, OA be the radius and AB be the chord, as shown in the figure.

The angle subtended by the chords at a point on the minor arc and also at a point on the major arc is 30o.

In ∆AOB,

Radius OA = Chord AB.

$\n \n \u21d2AO=OB=BA\n $ So, ∆AOB is an equilateral triangle.

$\n \n \u2220AOB=\n \n 60\n \xb0\n \n \n $

By degree measure theorem,

$\n \n \n \n \n \n \u2220AOB\n \n \n \n =2\u2220ACB\n \n \n \n \n \n \n \n 60\n \xb0\n \n \n \n \n \n =2\u2220ACB\n \n \n \n \n \n \u2220ACB\n \n \n \n =\n \n \n \n 60\n \xb0\n \n \n 2\n \n \n \n \n \n \n \n \u2220ACB\n \n \n \n =\n \n 30\n \xb0\n \n \n \n \n \n \n $

Opposite angles of cyclic quadrilateral are supplementary.

$\n \n \n \n \n \n \u2220ACB+\u2220ADB\n \n \n \n =\n \n 180\n \xb0\n \n \n \n \n \n \n \n \n \n 30\n \xb0\n \n +\u2220ADB\n \n \n \n =\n \n 180\n \xb0\n \n \n \n \n \n \n \n \u2220ADB\n \n \n \n =\n \n 180\n \xb0\n \n \u2212\n \n 30\n \xb0\n \n \n \n \n \n \n \n \u2220ADB\n \n \n \n =\n \n 150\n \xb0\n \n \n \n \n \n \n $

Therefore, angle by chord AB at minor arc = 150°.

Angle by chord AB at major arc = 30°.

Hence, the required answer is 30o.