Resposta :
senθ = 3/5
(sen θ.)^2 + (cos θ.)^2 = 1
(cos θ.)^2 = 1 - (3/5)^2
(cos θ.)^2 = 1 - 9/25 => (cos θ.)^2 = (25-9)/25 => (cos θ.)^2 = 16/25 => cos θ = 4/5
tg θ = senθ / cos θ => tg θ = 3/5 / 4/5 => tg θ = 3/4
cotg θ = 1/tg θ => cotg θ = 1/ 3/4 => cotg θ = 4/3
cossec θ = 1/sen θ => cossec θ = 1/ 3/5 => cossec θ = 5/3
sec θ = 1/cos θ => sec θ = 1/ 5/4 => sec θ = 5/4
Sendo sen B = 3/5
3 = cateto oposto
5 = hipotenusa
c^2 = 5^2 - 3^2 = 25 - 9 = 16
4 = cateto adjacente
cos B = 4/5
tag B = 3/4
cotg B = 1/tag = 4/3
sec B = 1/cos B = 5/5
cosc B = 1/senB = 5/3