[tex]A)sen30\°=\frac{1}{2}\\\frac{1}{2}=\frac{5}{x}\\x=5*2\\\\x=10\\\\B)sen60\°=\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}=\frac{5\sqrt3}{x}\\\sqrt3x=10\sqrt3\\x=\frac{10\sqrt3}{\sqrt3}\\\\x=10\\\\C)sen45\°=\frac{\sqrt2}{2}\\\frac{\sqrt2}{2}=\frac{x}{3\sqrt2}\\2x=3\sqrt2*\sqrt2\\2x=3*2\\2x=6\\x=6/2\\\\x=3\\\\D)tg60\°=\sqrt3\\\sqrt3=\frac{8\sqrt2}{x}\\sqrt3x=\frac{8\sqrt2}{x}\\\sqrt3x=8\sqrt2\\x=\frac{8\sqrt2}{\sqrt3}\\x=\frac{8\sqrt2*\sqrt3}{\sqrt3*\sqrt3}\\\\x=\frac{8\sqrt6}{3}[/tex]
boa noite.. espero ter ajudado.. marque a melhor resposta!! ;-)
