AB is the diameter of a circle centered at a point o BC chord is known that the angle AOC is 2 times greater than the angle

AB is the diameter of a circle centered at a point o BC chord is known that the angle AOC is 2 times greater than the angle AB find the angles AOC and COB.

We need to start with what we have. We have a circle inside which a triangle is built. AB – diameter. BC and CA chords, CO – radius. From this we start.
From the 6th grade, we know that the diameter is the radius multiplied by 2. That is, conventionally, the AB diameter can be divided into two equal segments (radii) BO and AO. If we connect point A to point C, we get a triangle. Thus, we need to find the angle of the OWL. And to get it, draw one more CO radius. (from point C to middle O) We get two small triangles.
It can be seen from the drawing that CO together with AO and OB are radii. And in a circle it is impossible to draw two different radii. Hence, AO = OB = CO by condition. Thus, we get two isosceles triangles AOC and BOS with a common CO side.
By condition, we know that the AOC angle is twice the ABC angle. (the angle is not fully spelled out in the problem, but there is no other option).
Let’s designate the AOC angle as 2x (since it is twice the unknown angle). The AOS triangle itself is isosceles. This means that the other two angles unknown to us are equal. Knowing that the sum of the angles of the triangle = 180, we find the remaining angles in the AOC triangle. 2x + x + x = 180. So 4x = 180. x = 45.
We got that the angles of the SAO and ASO are 45 degrees. The AOC angle was 2x, which means it = 90. The AOC and COB angles are adjacent. One of them = 90 degrees. An angle adjacent to a right angle is a right angle – recall the topic of adjacent corners. So we have solved the problem.
Answer: AOC angle = 90 degrees. OWL = 90 degrees. (adjacent)