Ta có: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) \( \Rightarrow {\sin ^2}\alpha + \dfrac{2}{{16}} = 1\) \( \Rightarrow {\sin ^2}\alpha = \dfrac{{14}}{{16}}\) \( \Rightarrow \sin \alpha = \dfrac{{\sqrt {14} }}{4}\) vì trong khoàng \(\left( {0;{{180}^0}} \right)\) thì \(\sin \alpha > 0\).
Suy ra \(\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}\)\( = \left( { - \dfrac{{\sqrt 2 }}{4}} \right):\dfrac{{\sqrt {14} }}{4} = - \sqrt 7 \).
