1. **State the problem:** We need to find the measure of angle RTS in triangle RTS inscribed in a circle. 2. **Given information:** - Angle at Q (inside the circle) is $3x + 57^\circ$. - Angle at T is $3x$. 3. **Important rule:** The sum of the interior angles of a triangle is $180^\circ$. 4. **Set up the equation:** $$\text{Angle } R + \text{Angle } T + \text{Angle } S = 180^\circ$$ Here, angle RTS is angle $R$, angle at T is $3x$, and angle at Q is $3x + 57^\circ$ which corresponds to angle $S$. So, $$m\angle RTS + 3x + (3x + 57) = 180$$ 5. **Simplify the equation:** $$m\angle RTS + 3x + 3x + 57 = 180$$ $$m\angle RTS + 6x + 57 = 180$$ 6. **Express $m\angle RTS$ in terms of $x$:** $$m\angle RTS = 180 - 6x - 57$$ $$m\angle RTS = 123 - 6x$$ 7. **Find $x$ using the fact that angle at Q is an exterior angle to triangle RTS:** Since angle at Q is $3x + 57^\circ$ and it equals the sum of the opposite interior angles $m\angle RTS + 3x$, we have: $$3x + 57 = m\angle RTS + 3x$$ 8. **Subtract $3x$ from both sides:** $$\cancel{3x} + 57 = m\angle RTS + \cancel{3x}$$ $$57 = m\angle RTS$$ 9. **Therefore,** $$m\angle RTS = 57^\circ$$ 10. **Final answer:** $$\boxed{57^\circ}$$