$\begin{array}{l}
\dfrac{\pi }{2} < \alpha < \pi \Rightarrow \sin \alpha > 0,\cos \alpha < 0\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1 \Rightarrow {\sin ^2}\alpha = 1 - {\left( {\dfrac{{ - 3}}{5}} \right)^2} = \dfrac{{16}}{{25}} \Rightarrow \sin \alpha = \dfrac{4}{5}
\end{array}$
