What is the Measure of Angle BAC in This Circle Geometry Problem?

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In the given geometry problem, point A lies on a circle centered at point O, and O is also on a circle centered at A. The angle BAC is determined to be greater than 90 degrees, and points B and C are identified as the intersection points of the two circles. Both circles have equal radii, leading to the conclusion that triangle AOB is equilateral, making all angles equal. The angle BAC is then expressed as the sum of angles BAO and OAC, which are derived from the properties of the equilateral triangles formed. The problem encourages further exploration of the angle's measure as an exercise for the student.
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In the euclidean plane, point A is on a circle centered at point O and point O is on a circle centered at point A. What is measure of angel BAC?

So I drew a picture, and it seems that BAC is going to definitely be greater than 90 degree's. From there I am confused on what to do next. Anyone have some advice?
 
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What is B and C?
 
I assume that B and C are the two common points on the circles.

Now, the radii of both circles are the same (why?). Now remember elementary geometry. Draw the circle around A using a compass. Put the point of the compass at point O and mark the point B (without changing the span of the compass). Now, the distance AO=r and AB=r (they are both on the circle around A). In addition, OB=r (the compass gave the distance). Thus, in the triangle AOB, all sides are equal, an therefore all angles are equal (the size of the angle is left as an exercise for the student).
Now, assume that you did not just mark point B, but let the compass draw a complete circle. In addition to point B you would also have a point C where the circles intersect. Then everything we just said about the triangle AOB is also valid for the triangle AOC. Now, the angle BAC is the sum of the angles BAO and OAC.
 

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