Point $P$ is inside equilateral $△ABC$. Points $Q$, $R$, and $S$ are the feet of the perpendiculars from $P$ to $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$, respectively. The side length of $△ABC$ is $10\sqrt{3}$. If $PQ=4$ and $PR=5$, what is $PS$?