1. **Stating the problem:** We have a circle with center O and points P, Q, R, S on the circumference forming quadrilateral PQRS. Outside the circle, triangle PST is connected with point T. Angles given are 60° at O, 70° and $y$ in triangle PST. We need to find the value of angle $y$. 2. **Understanding the geometry:** The 60° angle at O is likely the central angle subtending an arc of the circle. The 70° and $y$ angles are in triangle PST, which shares a side with the quadrilateral. 3. **Key rule:** The sum of angles in any triangle is 180°. 4. **Applying the rule to triangle PST:** $$70^\circ + y + \text{third angle} = 180^\circ$$ 5. **Using the given 60° angle at O:** Since 60° is a central angle, the arc it subtends is also 60°. The inscribed angle subtending the same arc is half of the central angle, so it is 30°. 6. **Assuming the third angle in triangle PST corresponds to this inscribed angle:** $$\text{third angle} = 30^\circ$$ 7. **Calculate $y$:** $$70^\circ + y + 30^\circ = 180^\circ$$ $$y + 100^\circ = 180^\circ$$ $$y = 180^\circ - 100^\circ$$ $$y = 80^\circ$$ **Final answer:** $$y = 80^\circ$$