In the given figure, AB is a diameter and AC is a chord of the circle and the tangent at C intersects AB produced in D. If ∠BAC=30∘, then ∠CDB=
30∘
∠ACB=90∘ (angles in the semicircle)
∠BCD=30∘ ( angles in the alternate segment are equal)
∠ACD=∠ACB+∠BCD
=90∘+30∘=120∘
∠CDB=180∘−(30∘+120∘) .....(sum of angles in a triangle)
=180∘−150∘=30∘