The principle you need to remember here is that the angles opposite equal sides are also equal.

It is given that AB = OC (radius)
We also know that OB is also the radius.
Therefore OB = OC
So AB = OB =OC

1:
Let's assume angle BAO = X (angle opposite OB)
Therefore angle BOA is also X (angle opposite AB)
Given angle COD = 60.
Angle BOC = 180 - 60 - X = 120 -X

We know that angle ABO = 180 - X- X= 180-2X
Angle OBC = 180 -(180-2X) = 2X
Using the same above principle angles opposite OB and OC are equal.
Angle OCB = 2X
Angle BOC = 180 - 2X - 2X= 180 - 4X

equate the two terms 120 -X = 180 -4X
X = 20 . So 1 is sufficient

Similar to this you can use 2 i.e. angle BCO to solve for X.

Therefore the answer is D

ok, this is a tough one.

first of all, we know that OC=OB since they are both the radius. not only that, but OB is shared by both triangle BAO and BCO. This is important because automatically, this tells us that angle BAO = angle BCO since they both face the same side. This point is key to answer the questions.

So for statement (1), given that angle COD is 60, angle COA must be 120 (180-60). Looking at triangle COA, we can figure out the other two angles (BAO and BCO) with the following formula, 120 +2x=180. 2x represents the fact that angle BAO must be equal to angle BCO as they both share the same side OB. So this is sufficient as we can conclude that angle BAO is 30.

For statement (2), again, because angle BCO and BAO are the same, this is pretty easy to answer. Angle BAO will also be 40.

Hope that helps. Good question!

Thanks. After exam when i looked at it i could figure it out. During exam the complicated figure puzzled me a little and it took more than 2 minutes. so i had move on.