1. **Stating the problem:** We are given a circle with center O and points A, B, C on the circle. The triangle OAB is formed such that $\angle OAB = 20^\circ$ and $\angle OCB = 50^\circ$. 2. **Understanding the problem:** Since O is the center of the circle, segments OA and OB are radii and thus equal in length. 3. **Key properties:** - Triangle OAB is isosceles with $OA = OB$. - The base angles of an isosceles triangle are equal. 4. **Find $\angle OBA$:** Since $\angle OAB = 20^\circ$, then $\angle OBA = 20^\circ$. 5. **Find $\angle AOB$:** Sum of angles in triangle OAB is $180^\circ$, so $$\angle AOB = 180^\circ - 20^\circ - 20^\circ = 140^\circ.$$ 6. **Relate $\angle AOB$ to $\angle ACB$:** The angle at the center $\angle AOB$ subtends the same arc AB as the angle at the circumference $\angle ACB$. 7. **Circle theorem:** The angle at the center is twice the angle at the circumference subtending the same arc, so $$\angle AOB = 2 \times \angle ACB.$$ 8. **Calculate $\angle ACB$:** $$\angle ACB = \frac{\angle AOB}{2} = \frac{140^\circ}{2} = 70^\circ.$$ 9. **Check given options:** The correct answer is (b) 70°. **Final answer:** $70^\circ$