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 = $\frac{AOC}{2} = \frac{120}{2} = 60^{\circ}$ (Angle subtended by the chord at the circle is half that subtended at the centre)Thus, in triangle ABC∠CAB + ∠CBA + ∠ACO + ∠OCB= 18060 + 60 + 30 + ∠OCB = 180∠OCB = 30°