In the given diagram O is the center of the circle and CD is a tangent. ∠CAB and ∠ACD are supplementary to each other ∠OAC = 30∘. Find the value of ∠OCB:
30∘
Since, OA = OC (Radii)
∠OAC = ∠OCA = 30∘
Therefore
∠AOC = 180∘ - ∠OAC - ∠OCA
= 180∘ – 30∘ – 30∘ = 120∘
∠ ABC = 12 ∠AOC
[The angle at the centre is twice the angle at the circumference.]
= 12×120∘ = 60∘
∠CAB and ∠ACD are supplementary ⇒ AB is parallel to CD
∠ABC = ∠BCD (Alternate interior angle)
∠BCD=60∘
From the figure,
∠ACB=∠ACO+∠OCB⇒60°=30°+∠OCB⇒∠OCB=60°−30°=30°