In the given figure, O is the center of the circle. If AB = OC, then what is the ratio of ∠CAO to the ∠DOC ?

1 : 4 2 : 7 2 : 5 1 : 3

1 : 3

OC = AB OC = OB  =  r ∠OBC = θ + θ = 2θ    (exterior angle) ∠OBC = ∠OCB = 2θ    (angle opposite to equal side area) ∠BOC = 180° - 4θ ∠DOC = 180° - (180° - 4θ + θ) = 3θ \(\frac{∠CAO}{∠DOC}\) = \(\frac{θ}{3θ}\) = \(\frac{1}{3}\)