Воспользуемся основным тригонометрическим тождеством: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\). Тогда:
\({\left( {-\dfrac{5}{{\sqrt {26} }}} \right)^2} + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\alpha = \dfrac{1}{{26}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos \alpha = \dfrac{1}{{\sqrt {26} }},\,\,\,}\\{\cos \alpha = -\dfrac{1}{{\sqrt {26} }}.}\end{array}} \right.\)
Так как \(\alpha \, \in \,\left( {\pi ;\dfrac{{3\pi }}{2}} \right)\) (III четверть), то \(\cos \alpha < 0\), то есть \(\cos \alpha =- \dfrac{1}{{\sqrt {26} }}.\)
Тогда: \({\rm{tg}}\alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = -\dfrac{5}{{\sqrt {26} }}:\left( {-\dfrac{1}{{\sqrt {26} }}} \right) = \dfrac{5}{{\sqrt {26} }} \cdot \dfrac{{\sqrt {26} }}{1} = 5.\)
Ответ: \(5.\)