$\triangle ABC$ is an acute angled triangle. Perpendiculars drawn from its vertices on the opposite sides are $AD$, $BE$ and $CF$. The line parallel to $ DF$ through $E$ meets $BC$ at $Y$ and $BA$ at $X$. $DF$ and $CA$ meet at $Z$. Circumcircle of $XYZ$ meets $AC$ at $S$. Given, $\angle B=33 ^\circ.$ find the angle $\angle FSD $ with proof.