Consider triangle COB has 2 equal sides, therefore two equal angles namely OBC and OCB let them both be X
Now triangle ABO will have two sides equal and hence two angles also namely BAO and BOA let them be Y.

Statement 1 :

COD= 60, we can write Y + BOC + 60 =180, Y+BOC = 120
from the stem we can figure out angle BOC in terms of X, BOC+2X=180, therefore BOC = 180-2X
replacing in the first equation we get: Y+180-2X=120 , 2X-Y = 60

Now we know angle ABO is external angle for triangle BOC, therefore ABO = X+BOC
ABO = X + 180 - 2X, ABO = 180 -X
ABO from triangle AOB can be written as : ABO=180-2Y, therefore equating both ABO we get: 180-2Y = 180-X, therefore X = 2Y.

Now we have two equations with X and Y you can solve for BOTH, Hence Statement 1 sufficient.

Given : AB = OC
We know, OB = OC = Radius of the circle.
We have two iscosceles triangles ABO and OBC.

Let, ang.BAO = ang.BOA = x (opposite angles of a iso. triangle)
ang.OBC =ang.OCB=2x (external angle + opposite angles of a iso. triangle)

1. ang.COD =60, therefore ang. COA = 120
x+2x+120 = 180
therefore ang. BAO = x = 20

2. ang. BCO = 40 = 2x
therefore ang. BAO = x = 20


Hope this helps.

Thanks guys!