Question

Daca ∠ ∈ (0° ; 90°) si cos∠=$\frac{3}{5}$ , aflati sin∠ si tg∠.

Answer 1

sin²x+cos²x=1
sin²x=1-cos²x
sinx=±√(1-cos²x) ,dar x∈(0° ; 90°) ⇒sinx=√(1-cos²x)
$sinx= \sqrt{1-( \frac{3}{5}) ^{2}} = \sqrt{1- \frac{9}{25}}= \sqrt{\frac{16}{25} } = \frac{4}{5}\\ \\ \\ tgx= \frac{sinx}{cosx}= \frac{ \frac{3}{5} }{ \frac{4}{5} } = \frac{3}{4}$