1. **State the problem:** We have an isosceles triangle PQR with sides RP = RQ and angles at P and Q given as $3x + 10$ and $x + 52$ degrees respectively. We need to find the value of angle $y$ at vertex R. 2. **Recall the properties:** In an isosceles triangle, the angles opposite the equal sides are equal. Since $RP = RQ$, the angles at P and Q are equal. 3. **Set up the equation:** Since angles at P and Q are equal, $$3x + 10 = x + 52$$ 4. **Solve for $x$:** $$3x + 10 = x + 52$$ $$3x - x = 52 - 10$$ $$2x = 42$$ $$x = 21$$ 5. **Find the angles at P and Q:** $$3x + 10 = 3(21) + 10 = 63 + 10 = 73$$ $$x + 52 = 21 + 52 = 73$$ 6. **Use the triangle angle sum rule:** The sum of angles in a triangle is $180$ degrees. $$y + 73 + 73 = 180$$ $$y + 146 = 180$$ $$y = 180 - 146 = 34$$ **Final answer:** $$y = 34$$ degrees.