In Fig. 16.196, if ∠AOB = 80° and ∠ABC=30°, then find ∠CAO.
2 ∠OAB = 100°
∠OAB = 50°
Therefore,
∠OAB = ∠OBA = 50°
∠AOB = 2 ∠BCA (Angle subtended by any point on circle)
80° = 2 ∠BCA
∠BCA = 40°
Now,
In triangle ABC
∠A + ∠B + ∠C = 180°
∠A + 30o + 40o = 180°
∠A = 110°
∠CAB = ∠CAO + ∠OAB
110° = ∠CAO + 50°
∠CAO = 60°