\(\sin\alpha=0,6=\dfrac{3}{5}\\ \sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\\ \Leftrightarrow\cos\alpha=\dfrac{4}{5}\\ \tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\left(=0,75\right)\)

