Question

Sendo sen β = 2/5 encontre os valores de:
a) Cos β =
b) tg β =
c) sec β

Answer 1

Resposta:

a) cos β = raiz(21)/5

b) tg β = 2.raiz(21)/21

c) sec β = 5.raiz(21)/21

Explicação passo-a-passo:

Temos que:

a)

sen^2 β + cos^2 β = 1

(2/5)^2 + cos^2 β = 1

cos^2 β = 1 - 4/25

cos^2 β = (25-4)/25

cos^2 β = 21/25

cos β = raiz(21/25)

cos β = raiz(21)/5

b)

tg β = sen β / cos β

tg β = 2/5 / raiz(21)/5

tg β = 2/raiz(21). raiz(21)/raiz(21)

tg β = 2.raiz(21)/21

c)

sec β = 1/cos β

sec β = 5/raiz(21). raiz(21)/raiz(21)

sec β = 5.raiz(21)/21

Blz?

Abs :)