Câu 1:
Ta có: \(\cos\left(90^0-\alpha\right)=\sin\alpha\)
\(\Leftrightarrow\sin\alpha=1:\sqrt{\dfrac{1^2+2^2}{1}}=1:\sqrt{5}=\dfrac{\sqrt{5}}{5}\)
Câu 2:
a) \(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\)
\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)