Question

Calcule a medida do lado ABC do triângulo abaixo
AB= 2√3
BC= x
AC= 2√7

Answer 1

${(2\sqrt{7})}^{2}={(2\sqrt{3})}^{2}+{x}^{2}-2.2\sqrt{3}.x}.\cos(30\degree)$

$28=12+{x}^{2}-2.\cancel{2}\sqrt{3}.x.\dfrac{\sqrt{3}}{\cancel{2}}$

$28=12+{x}^{2}-6x\\{x}^{2}-6x+12-28=0[/tex</p><p>[tex]{x}^{2}-6x-16=0\\{x}^{2}+(2-8)x+2.(-8)=0\rightarrow\,(x+2)(x-8)=0$

$x-8=0\\\bf{x=8}$