Question

Dado triângulo abaixo, calcule Tg x.

Answer 1

Seja $y$ o angulo oposto ao lado do triangulo menor de comprimento 3 , sabemos que

$tg(x+y)=\dfrac{7}{10} ~~ (i)\\\\tg(y)=\dfrac{3}{10} ~~(ii)$

De $(i)$ temos

$tg(x+y)=\dfrac{tg (x) +tg (y)}{1-tg(x)\cdot tg(y)}=\dfrac{7}{10}~~(iii)$

Substituindo $(ii)$ em$(iii)$

$\dfrac{ \dfrac{3}{10} +tg (x)}{1-\dfrac{3}{10}\cdot tg(x)}=\dfrac{7}{10}\Rightarrow \\\\\\\dfrac{10tg (x)+3}{10-3tg(x)}=\dfrac{7}{10}\\\\100tg(x)+30=70-21tg(x)\\\\121tg(x)=40\Rightarrow \boxed{tg(x)=\dfrac{40}{121}}$

Answer 2

Resposta:

tgx = 40/121

Explicação passo-a-passo:

tgy = 3/10

tg(x + y) = 7/10

$\frac{tgx+tgy}{1-tgx.tgy}=\frac{7}{10}\\\\\frac{tgx+\frac{3}{10}}{1-tgx.\frac{3}{10}}=\frac{7}{10}\\\\\frac{\frac{10tgx+3}{10}}{\frac{10-3tgx}{10}}=\frac{7}{10}\\\\\frac{10tgx+3}{10-3tgx}=\frac{7}{10}\\\\100tgx+30=70-21tgx\\\\121tgx=40\\\\tgx=\frac{40}{121}$