Воспользуемся формулой: \(1 + {\rm{t}}{{\rm{g}}^2}\alpha = \dfrac{1}{{{{\cos }^2}\alpha }}.\)
\(1 + {\left( {\dfrac{{12}}{5}} \right)^2} = \dfrac{1}{{{{\cos }^2}\alpha }}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\alpha = \dfrac{{25}}{{169}}.\)
Используя основное тригонометрическое тождество, получим:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha + \dfrac{{25}}{{169}} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha = \dfrac{{144}}{{169}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \alpha = \dfrac{{12}}{{13}},\,\,\,}\\{\sin \alpha = -\dfrac{{12}}{{13}}.}\end{array}} \right.\)
Так как \(\alpha \, \in \,\left( {\pi ;\dfrac{{3\pi }}{2}} \right)\) (III четверть), то \(\sin \alpha < 0\), то есть \(\sin \alpha = -\dfrac{{12}}{{13}}.\)
Тогда: \(169\sin \alpha = 169 \cdot \left( {-\dfrac{{12}}{{13}}} \right) = -156.\)
Ответ: \(-156.\)