SSC CHSL Model Paper 6

∠OCD = 90° (CD is a tangent and OC is radius at the point of tangency)
∠OAC = ∠OCA = 30° (Since, they are opposite angles to equal line segments, OA = OC)
∠ACD = ∠ACO + ∠OCD
= 30° + 90°
= 120°
∠BAC = 180° - 120° = 60° (Given ∠CAB and ∠ACD are supplementary)
In triangle AOC
∠OAC + ∠ OCA + ∠AOC = 180° (Sum of angles of a triangle)
∠AOC= 120°
Hence, ∠ABC = $\frac{\mathrm{AOC}}{2}=\frac{120}{2}=60°$ (Angle subtended by the chord at the circle is half that subtended at the centre)
Thus, in triangle ABC
∠CAB + ∠CBA + ∠ACO + ∠OCB= 180
60 + 60 + 30 + ∠OCB = 180
∠OCB = 30°