In a circle with center O, diameters are drawn so that the BC chord is equal to the radius. Find the corner of the AOB.

Let AC and ВD be the diameters of the circle.
In the triangle ВOС, OB = OС = R.
By condition, BC is equal to the radius, then in the ВOС triangle, OB = OС = BC = R, then the ВOС triangle is equilateral and all of its internal angles are 60.
The sought angle AOB is adjacent to the angle BOC, the sum of which is 1800.
Then the angle AOB = (180 – BOS) = (180 – 60) = 120.
Answer: The angle AOB is 120.