para o triângulo apresentado, determine a distância de x.

Resposta:
[tex]\textsf{Leia abaixo}[/tex]
Explicação passo a passo:
[tex]\mathsf{a^2 = b^2 + c^2 -2.b.c.cos\:\Theta }[/tex]
[tex]\mathsf{(4\sqrt{2})^2 = 6^2 + x^2 -2.6.x.cos\:60\textdegree }[/tex]
[tex]\mathsf{32 = 36 + x^2 -2.6.x.\dfrac{1}{2} }[/tex]
[tex]\mathsf{x^2 - 6x + 4 = 0 }[/tex]
[tex]\mathsf{\Delta = b^2 - 4.a.c}[/tex]
[tex]\mathsf{\Delta = (-6)^2 - 4.1.4}[/tex]
[tex]\mathsf{\Delta = 36 - 16}[/tex]
[tex]\mathsf{\Delta = 20}[/tex]
[tex]\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{6 \pm \sqrt{20}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{6 + 2\sqrt{5}}{2} = 3 + \sqrt{5}}\\\\\mathsf{x'' = \dfrac{3 - 2\sqrt{5}}{2} = 3 - \sqrt{5}}\end{cases}}[/tex]
[tex]\boxed{\boxed{\mathsf{S = \{3 + \sqrt{5};\:3 - \sqrt{5}\}}}}[/tex]