Question

Answer 1

\(sin^2a+cos^2a=1\\ =>cos^2a=1-sin^2a=1-\left(\dfrac{4}{5}\right)^2\\ =\dfrac{9}{25}\\ =>cosa=\dfrac{3}{5}\\ tana=\dfrac{sina}{cosa}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}\)

Answer 2

Ta có: `sin^2 \alpha+cos^2 \alpha=1`

`<=>(4/5)^2+cos^2 \alpha=1`

`<=>cos \alpha=[+-3]/5`

  Mà `\alpha` là góc nhọn

 `=>cos \alpha=3/5`

Có: `tan \alpha=[sin \alpha]/[cos \alpha]=[4/5]/[3/5]=4/3`