Question

Determine o valor de x no triângulo abaixo

Answer 1

Resposta:

Explicação passo-a-passo:

Lei dos senos

$\frac{x}{sen(\alpha )} =\frac{y}{sen(\beta) } \\\\\frac{6}{sen(45^{\circ})} =\frac{a}{sen(60^{\circ})} \\\\\frac{6}{\frac{\sqrt{2} }{2} } =\frac{a}{\frac{\sqrt{3} }{2} } \\\\\frac{12}{\sqrt{2} } =\frac{2a}{\sqrt{3} } \\\\12\sqrt{3} =2\sqrt{2} a\\\\a=\frac{12\sqrt{3} }{2\sqrt{2} } \\\\a=\frac{12\sqrt{3}\times\sqrt{2} }{2\sqrt{2} \times\sqrt{2} } \\\\a=\frac{12\sqrt{6} }{4}\\ \\\boxed{\boxed{a=3\sqrt{6} \:cm}}$