I'm sure there is some property of congruent angles that I don't remember that would help me solve this one but I'll be darned if I know what it is.
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Nice question !
OB=OC = radius. So angles BCO and CBO are equal.
Since AB=OC, we have AB=OC=OB. So angles BAO and BOA are equal.
Also, exterior angle CBO = sum of interior angles = BOA + BAO
Let x = BAO. We have
BAO = x = BOA and CBO = BCO = 2x
1 - sufficient. COD = 60 implies COA = 120 = COB + BOA
COB = 180 - 4x (property of triangles)
COB + BOA = 180 - 4x + x = 180 - 3x = 120.
We can solve for x and hence get BAO
2 - sufficient. BCO = 2x. So, we can get x and hence angle BAO.
Hence D.
