Question

Considerando o triângulo e retângulo , determine :

Answer 1

Resposta:

Explicação passo a passo:

a)

$tg (\alpha ) = \frac{8}{8\sqrt{3} } = \frac{1}{\sqrt{3} }=\frac{\sqrt{3} }{3}$

Logo, $\alpha = 30^o$

b) Aplicando

$sen(\alpha ) = \frac{8}{AC}$

$sen(30) = sen(\alpha) = \frac{1}{2}$

$\frac{1}{2} = \frac{8}{AC}$

$AC = 16$

c)

sen 60 = $\frac{\sqrt{3} }{2\\}$

cos 60 = $\frac{1}{2}$

tg (60) = $\sqrt{3}$

d)

$sen \alpha =\frac{8}{16} = \frac{1}{2}$

$cos \alpha = \frac{8\sqrt{3} }{16} = \frac{\sqrt{3} }{2}$

$tg (\alpha ) = \frac{8}{8\sqrt{3} } = \frac{1}{\sqrt{3} }=\frac{\sqrt{3} }{3}$