1. **Stating the problem:** We have two separate geometry problems involving angles in circles. 2. **Problem 6a:** Find $\angle ACB$ given that it is a straight angle. 3. **Solution 6a:** Since $\angle ACB$ is given as 180°, it is a straight angle by definition. 4. **Problem 6b:** Given $\angle 1 = 50^\circ$, no calculation needed. 5. **Problem 8a and 8b:** Given a circle with center $O$, line $AB$ passing through $O$, and $\angle AOB = 80^\circ$. 6. **Step for 8a:** $\angle 1$ is an inscribed angle subtending the same arc as $\angle AOB$. 7. **Formula:** The inscribed angle theorem states: $$\angle 1 = \frac{1}{2} \times \angle AOB$$ 8. **Calculation:** $$\angle 1 = \frac{1}{2} \times 80^\circ = 40^\circ$$ 9. **Step for 8b:** Given two chords intersecting inside the circle creating angles $\angle 1$ and $\angle 2$, with arcs 160° and 140°. 10. **Formula:** The angle formed by two intersecting chords is half the sum of the measures of the arcs intercepted by the angle and its vertical angle: $$\angle 1 = \angle 2 = \frac{1}{2} (160^\circ + 140^\circ)$$ 11. **Calculation:** $$\angle 1 = \angle 2 = \frac{1}{2} \times 300^\circ = 150^\circ$$ **Final answers:** - 6a. $\angle ACB = 180^\circ$ - 6b. $\angle 1 = 50^\circ$ - 8a. $\angle 1 = 40^\circ$ - 8b. $\angle 2 = 150^\circ$