Question

Sen B e cotg B, sabendo que cos B=√6/5
Determine

Answer 1

Resposta:

$cos\beta =\frac{\sqrt{6}}{5}}$

Cos = cateto adjacente/hipotenusa

hipotenusa = 5

Descobrindo seno de beta:

$5^2=(\sqrt{6})^2+x^2$

$25=6+x^2$

$25-6=x^2$

$x=\sqrt{19$

$Sen\beta =\frac{\sqrt{19}}{5}$

Tangente = $tg\beta =\frac{sen\beta }{cos\beta }$

$tg\beta =\frac{\sqrt{19}}{5} \cdot \frac{5}{\sqrt{6}}$

$tg\beta =\frac{\sqrt19}{\sqrt6}$

$tg\beta =\sqrt{\frac{19}{6}}$

Assim temos que

$cotg\beta =\frac{1}{\sqrt{\frac{19}{6}}}$

racionalizando o denominador teremos

$cotg \beta =\sqrt{\frac{6}{19}}$

Espero ter ajudado