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To determine the size of \(\angle OBC\), we can use the following approach:

1. Since \(O\) is the center of the circle, \(\overline{OA}\), \(\overline{OB}\), and \(\overline{OC}\) are radii of the circle.
2. We are given that \(\angle AOC = 120^\circ\), which is the angle at the center.

Now, let's apply the **angle subtended by the same arc theorem**, which states that the angle subtended by an arc at the center of the circle is **twice** the angle subtended by the same arc at the circumference.

- Arc \(AC\) subtends \(\angle AOC\) at the center, which is \(120^\circ\).
- Therefore, the angle subtended by arc \(AC\) at point \(B\) on the circumference, \(\angle ABC\), will be half of \(\angle AOC\).

\[
\angle ABC = \frac{1}{2} \times 120^\circ = 60^\circ
\]

Thus, \(\angle OBC\) is **60°**.

Let me know if you need more details or have further questions!

Here are 5 related questions for deeper understanding:
1. What is the measure of \(\angle ABC\) in this circle?
2. How can you prove that \(\triangle OBC\) is isosceles?
3. What is the total sum of the interior angles in \(\triangle OBC\)?
4. How would you calculate the length of \(\overline{OC}\) if the radius of the circle is known?
5. How does the property of cyclic quadrilaterals apply to this diagram?

**Tip:** Always remember that angles at the center of a circle are twice the angles on the circumference subtended by the same arc.

Learn how to calculate the size of angle OBC in a circle where the central angle AOC is given as 120°. This solution uses the angle subtended by the same arc theorem, which states that the angle at the center is twice the angle at the circumference.

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