Ta có sin α = \(\frac{4}{5}\)
⇒cos α = \(\sqrt{1-\sin^2\alpha}\)
=\(\sqrt{1-\left(\frac{4}{5}\right)^2}\)
=\(\sqrt{\frac{9}{25}}\)
=\(\frac{3}{5}\)
Ta có tan α = \(\frac{\sin\alpha}{\cos\alpha}\)
=\(\frac{\frac{4}{5}}{\frac{3}{5}}\)
=\(\frac{4}{3}\)