По условию \(\sin A = \dfrac{{2\sqrt 6 }}{5}.\) Для нахождения \(\cos A\) воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}A + {\cos ^2}A = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{\left( {\dfrac{{2\sqrt 6 }}{5}} \right)^2} + {\cos ^2}A = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,{\cos ^2}A = 1-\dfrac{{24}}{{25}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{\cos ^2}A = \dfrac{1}{{25}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\cos A = \dfrac{1}{5} = 0,2.\)
Ответ: 0,2.