Question

Dado o triângulo retângulo, calcule os valores de sem a e cos a​

Answer 1

$\large\boxed{\begin{array}{l}\rm x\longrightarrow cateto\,adjacente\,a\,\alpha\\\rm x^2+6^2=10^2\\\rm x^2+36=100\\\rm x^2=100-36\\\rm x^2=64\\\rm x=\sqrt{64}\\\rm x=8\\\rm sen(\alpha)=\dfrac{6\div2}{10\div2}=\dfrac{3}{5}\\\\\rm cos(\alpha)=\dfrac{8\div2}{10\div2}\\\\\rm cos(\alpha)=\dfrac{4}{5}\end{array}}$

Answer 2

Resposta:

A seguir.

Explicação passo a passo:

sena = 6/10 = 3/5

10² = 6² + x²

x² = 100 - 36 = 64

x = √64

x = 8

cosa = 8/10 = 4/5