Question

Calcule o valor de x no triângulo abaixo.

Answer 1

Explicação passo-a-passo:

Pela lei dos cossenos:

$\sf (\sqrt{109})^2=x^2+5^2-2\cdot x\cdot5\cdot cos~120^{\circ}$

$\sf (\sqrt{109})^2=x^2+5^2-2\cdot x\cdot5\cdot\left(-\dfrac{1}{2}\right)$

$\sf 109=x^2+25+5x$

$\sf x^2+5x+25-109=0$

$\sf x^2+5x-84=0$

$\sf \Delta=5^2-4\cdot1\cdot(-84)$

$\sf \Delta=25+336$

$\sf \Delta=361$

$\sf x=\dfrac{-5\pm\sqrt{361}}{2\cdot1}=\dfrac{-5\pm19}{2}$

$\sf x'=\dfrac{-5+19}{2}~\Rightarrow~x'=\dfrac{14}{2}~\Rightarrow~\red{x'=7}$

$\sf x"=\dfrac{-5-19}{2}~\Rightarrow~x"=\dfrac{-24}{2}~\Rightarrow~x"=-12$ (não serve)

Logo, $\sf x=7$