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condition I is sufficient (hard to explain without diagram, I'll post the diagram later), but angle BAO comes out to be 20

condition II is sufficient because:

Given that <BCO is 40, this means <CBO is also 40, since triangle BCO is isoceles (because the 2 radii are same).

Now <CBO and <ABO are supplementary, meaning they must add up to 180, so that leaves us <ABO = 140.

Finally, triangle ABO is also isoceles, because segment AB is the same as the radius. So that means each of the other 2 angles of triangle ABO is equal to 20

Statement 1: First of let us establish few things - BAO and BOC are iscoseles triangles. so angles BAO=BOA; OBC=OCB Given that COD=60. Assuming BOC=x BOA=180-x-60=120-x...1 also, OBC=(180-x)/2 => OBA = 180-(180-x)/2 => BOA=(180-OBA)/2= (180-(180-(180-x)/2)/2 = (180-x)/4...2 Equating 1 and 2 => 120-x=(180-x)/4 => 480-4x=180-x => x=100 => BAO=BOA=180-100-60=20 You really don't need to solve till the end in the exam, once you get a feel that you can solve for x you are pretty much done.

My answer is also D, similar to Sri's logic Triangles BOC & BOA are both isosceles triangles and hence have equal bases. Therefore you only need to determine one angle in triangle BOA. With statement 1, you can deduce angle BOA ( as triangle in the semi circle is a right angle trinangle; the other angle is provided by the statement and 180- both gives angle BOA). And since BOA=BAO, the statement is sufficient. Similarly with second statement you can get angle OBC, which will help deduce angle OBA. (180-angle OBA)/2 will be the solution.

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