Question

Calcule o valor de x em cada caso :

Answer 1

$(4\sqrt{2} )^2=6^2+x^2-2.6xcos60\\\\16.2=36+x^2-6x\\\\x^2-6x+4=0$

Δ = 36 - 16 = 20

$x'=\frac{6+2\sqrt{5} }{2} =3+\sqrt{5} \\\\x''=3-\sqrt{5}$