jmc

Let $C$ be the point where line segment $\overline{OT}$ intersects the circle.



Since $\angle ROB=3\angle SOT=3\angle SOC$, the measure of arc $RB$ is three times the measure of arc $SC$. We also have $$\angle RTB=\frac{\widehat{RB}-\widehat{SC}}{2}.$$ Letting the measure of $\widehat{SC}$ be $x$, we have $\angle RTB=(3x-x)/2=x$, so $x=28^{\circ }$. Therefore, we have $\widehat{RB}=3x=84^{\circ }$ and $\widehat{SC}=28^{\circ }$. Since $\widehat{BC}$ is a semicircle, we have $\widehat{RS}=180^{\circ }-\widehat{RB}-\widehat{SC}=180^{\circ }-84^{\circ }-28^{\circ }=\boxed{68^{\circ }}$.