In the following figure (not to scale), at the centre O if the chord AB subtends double the angle that is subtended by chord CD and the angle $$\angle$$AEB = 2$$\angle$$AOB, then $$\angle$$COD is equal to:

Given, the chord AB subtends double the angle that is subtended by chord CD at the centre O

$$=$$>  $$\angle$$AOB = 2$$\angle$$COD

Let $$\angle$$COD = a

$$=$$>  $$\angle$$AOB = 2a

$$\angle$$AEB = 2$$\angle$$AOB

$$=$$>  $$\angle$$AEB = 2(2a) = 4a

In quadrilateral OAEB,

$$\angle$$AOB + $$\angle$$OBE + $$\angle$$AEB + $$\angle$$OAE = 360$$^\circ$$

$$=$$>  2a + 90$$^\circ$$ + 4a + 90$$^\circ$$ = 360$$^\circ$$

$$=$$>  6a + 180$$^\circ$$ = 360$$^\circ$$

$$=$$>  6a = 180$$^\circ$$

$$=$$>  a = 30$$^\circ$$

$$=$$>  $$\angle$$COD = 30$$^\circ$$

Hence, the correct answer is Option A