Question

Calcule o valor de y no triângulo abaixo

Answer 1

$tg(75) = \frac{tg(30) + tg(45)}{1 - tg(30).tg(45)} \\ = \frac{ \frac{ {\sqrt{3} } }{3} + 1}{1 - \frac{ \sqrt{3} }{3}.1 }$

$= \frac{ \frac{ \sqrt{3} + 3 }{3} }{ \frac{ 3- \sqrt{3} }{3} } \\ = \frac{ \sqrt{3} + 3 }{3 - \sqrt{3} }$

$\frac{ (\sqrt{3} + 3) }{(3 - \sqrt{3}) } . \frac{(3 + \sqrt{3} )}{(3 + \sqrt{3} )} \\ = \frac{9 + 6 \sqrt{3 } + 3}{9 - 3} = \frac{12 + 6 \sqrt{3} }{6} \\ = 2 + \sqrt{3}$

$tg(75) = \frac{y}{5} \\ 2 + \sqrt{3} = \frac{y}{5} \\ y = 5(2 + \sqrt{3} )$