Question

Calcule a medida de AB no triângulo abaixo:

Answer 1

Resposta:

$\frac{sen45}{6\sqrt{2} } =\frac{sen30}{x}$

$\frac{\sqrt{2} }{2} :6\sqrt{2} =\frac{1}{2} :x$

$\frac{\sqrt{2} }{2} .\frac{1}{6\sqrt{2} } =\frac{1}{2} .\frac{1}{x}$

$\frac{\sqrt{2} }{12\sqrt{2} }=\frac{1}{2x}$

$2\sqrt{2} x=12\sqrt{2} \\\\x=\frac{12\sqrt{2} }{2\sqrt{2} } \\\\x=\frac{12}{2} \\\\x=6$