Find the angles A, B, and ACB.

A short explanation:

Step-by-step solution:

1. Step 1: The angle subtended by arc BC at the center O is ∠BOC. Since ∠BAC subtends arc BC at the circumference, ∠BOC = 2 * ∠BAC. We are given that ∠A = 48 degrees, so ∠BOC = 2 * 48° = 96°.
2. Step 2: In triangle BOC, OB = OC (radii), so it is an isosceles triangle. Therefore, ∠OBC = ∠OCB. The sum of angles in triangle BOC is 180°. So, ∠BOC + ∠OBC + ∠OCB = 180°. Substituting ∠BOC = 96°, we get 96° + 2 * ∠OCB = 180°. This gives 2 * ∠OCB = 180° - 96° = 84°. Thus, ∠OCB = 84° / 2 = 42°.
3. Step 3: Since ∠OBC = ∠OCB, we have ∠OBC = 42°.
4. Step 4: ∠A is given as 48°.
5. Step 5: Angle ∠B is the same as ∠OBC, which is 42°.
6. Step 6: Angle ∠ACB is the sum of ∠ACO and ∠OCB. We need to find ∠ACO. In triangle AOC, OA = OC (radii), so it is an isosceles triangle. Angle ∠AOC = 180° - ∠BOC = 180° - 96° = 84° (angles on a straight line if AB is a diameter, but it is not given). Let's find ∠AOC differently. Angle ∠ABC subtends arc AC. We don't know ∠ABC yet.
7. Alternative approach for Step 6: Angle ∠A = 48°. Angle ∠B = 42°. In triangle ABC, the sum of angles is 180°. So, ∠ACB = 180° - ∠A - ∠B = 180° - 48° - 42° = 180° - 90° = 90°.
8. Final check: If ∠ACB = 90°, then AB is a diameter. If AB is a diameter, then the center O lies on AB. In triangle AOC, OA=OC, so ∠OAC = ∠OCA = 48°. Then ∠AOC = 180 - (48+48) = 84°. In triangle BOC, OB=OC, so ∠OBC = ∠OCB. ∠AOC + ∠BOC = 180° (if AB is a diameter). So ∠BOC = 180 - 84 = 96°. Then ∠OBC = ∠OCB = (180-96)/2 = 84/2 = 42°. ∠A = 48°. ∠B = 42°. ∠ACB = ∠ACO + ∠OCB = 48° + 42° = 90°. This is consistent.

Answer: ∠A = 48°, ∠B = 42°, ∠ACB = 90°