Question

sabendo que sen =3/7,determine : cos(x) ,tg(x), cotg(x) ,cossec(x),sec(x)

Answer 1

Resposta:

h=7

oposto = 3

adjacente = 2 raiz de 10

${7}^{2} = {3}^{2} + {a}^{2} \\ 49 - 9 = {a}^{2} \\ a = \sqrt{40} \\ a = 2 \sqrt{10}$

cos x= adj/ h

$\frac{2 \sqrt{10} }{7}$

tg x = o/a

$\frac{3}{2 \sqrt{10} } \\ \frac{3 \times 2 \sqrt{10} }{40} \\ \frac{ 6\sqrt{10} }{40} \\ \frac{3 \sqrt{10} }{20}$

cotg x= a/O

$\frac{2 \sqrt{10} }{3}$

cossec x = h/O

$\frac{7}{3}$

sec x = h/ a

$\frac{7}{2 \sqrt{10} } \\ \frac{7 \times 2 \sqrt{10} }{40} \\ \frac{7 \sqrt{10} }{20}$