To solve the problem (assuming the typical setup where (AB) is the diameter of the circle with center (O), (\angle BAC = 30^\circ), and we need (\angle OCB)):

Step 1: Use Thales' Theorem

Since (AB) is the diameter, (\angle ACB = 90^\circ) (Thales' theorem: angle subtended by diameter is right angle).

Step 2: Calculate (\angle ABC)

In (\triangle ABC), sum of angles is (180^\circ):
(\angle ABC = 180^\circ - \angle BAC - \angle ACB = 180^\circ - 30^\circ - 90^\circ = 60^\circ).

Step 3: Analyze (\triangle OBC)

(OB = OC) (radii of the circle), so (\triangle OBC) is sceles.
(\angle OBC = \angle ABC = 60^\circ) (since (O) is the midpoint of (AB)).

Thus, (\triangle OBC) is equilateral (all angles (60^\circ)), so (\angle OCB = 60^\circ).

Answer: (\boxed{60})

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