Question

1) Dado o triângulo retângulo abaixo, calcule tan x.

Answer 1

Chamaremos o ângulo adjacente ao x (do lado do x) de β.

Temos que:

$tg \beta = \frac{CO}{CA} \\\\ tg \beta = \frac{4}{10} \\\\ \boxed{tg \beta = \frac{2}{5}}$

Agora podemos calcular a tangente do ângulo maior (x+β):

$tg(x+\beta) = \dfrac{tgx+tg\beta}{1-tgx \cdot tg\beta} \\\\\\ \dfrac{6+4}{10} = \dfrac{tgx+\dfrac{2}{5}}{1-tgx \cdot \dfrac{2}{5}} \\\\\\ tgx+\dfrac{2}{5} = 1-\dfrac{tgx \cdot 2}{5} \\\\\\ tgx+\dfrac{tgx \cdot 2}{5} = 1-\dfrac{2}{5} \\\\\\ MMC = 5$

$\dfrac{5 \cdot tgx+2 \cdot tgx}{5} = \dfrac{5-2}{5} \\\\\\ 7 \cdot tgx = 3 \\\\\ \therefore \boxed{\boxed{tgx = \frac{3}{7}}}$