1. **State the problem:** We have a quadrilateral NOPQ inscribed in circle R with angles at N, O, P, and Q given as 52°, 74°, (x-89)°, and (2y-38)° respectively. We need to find the values of $x$ and $y$. 2. **Key property:** Opposite angles of a cyclic quadrilateral sum to 180°. 3. **Set up equations:** - Angles at N and P are opposite, so: $$52 + (2y - 38) = 180$$ - Angles at O and Q are opposite, so: $$74 + (x - 89) = 180$$ 4. **Solve first equation:** $$52 + 2y - 38 = 180$$ $$\cancel{52} + 2y - \cancel{38} = 180$$ $$14 + 2y = 180$$ $$2y = 180 - 14$$ $$2y = 166$$ $$y = \frac{166}{2} = 83$$ 5. **Solve second equation:** $$74 + x - 89 = 180$$ $$\cancel{74} + x - \cancel{89} = 180$$ $$x - 15 = 180$$ $$x = 180 + 15 = 195$$ 6. **Final answers:** $$x = 195$$ $$y = 83$$