AB is a chord in a circle with centre O. AB is produced to C such that BC is equal to the radius of the circle. C is joined to O and produced to meet the circle at D. If $$\angle ACD = 32^\circ$$, then the measure of $$\angle AOD$$ is .........

BC = OB (Given)

$$\angle BCO = \angle BOC = 32\degree = \angle ACD$$

$$\angle OBC = 180 - \angle BCO - \angle BOC$$ = 180 - 32 - 32 = 116

$$\angle BOD = \angle OBC + \angle OCB$$ = 116 + 32 = 148

In triangle AOB - 

$$\angle ABO = 180 - \angle$$ OBC = 180 - 116 = 64

$$\angle ABO = \angle OA$$ = 64 

($$\because$$OA = OB)

$$\angle AOB = 180 - \angle ABO - \angle$$ OAB = 180 - 64 -64 = 52

$$\angle AOD = \angle BOD - \angle AOB = 148 - 52 = 96\degree$$