Let $\angle TSU =x$. In $\Delta STU$, we have
$\begin{array}{rcll}
ST & = & TU & \text{(given)} \\
\angle TUS & = & \angle TSU & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle TUS & = & x
\end{array}$
In $\Delta SUV$,
$\begin{array}{rcll}
\angle RSU & = & \angle SUV +\angle SVU & \text{(ext. $\angle$ of $\Delta$)} \\
\angle RSU & = & x+48^\circ
\end{array}$
In $\Delta SUW$,
$\begin{array}{rcll}
\angle RUS & = & \angle SWU +\angle USW & \text{(ext. $\angle$ of $\Delta$)} \\
\angle RUS & = & 32^\circ +x
\end{array}$
Since $RSTU$ is a cyclic quadrilateral, we have
$\begin{array}{rcll}
\angle RUT +\angle RST & = & 180^\circ & \text{(opp. $\angle$s, cyclic quad.)} \\
\angle RUS+\angle SUT +\angle RSU +\angle UST & = & 180^\circ \\
32^\circ +x+x+x+48^\circ+x & = & 180^\circ \\
4x & = & 100^\circ \\
x & = & 25^\circ
\end{array}$
Hence, we have
$\begin{array}{rcl}
\angle RSU & = & 25^\circ+48^\circ \\
\angle RSU & = & 73^\circ
\end{array}$
