Question

determine amedida do lado ab dos triangulos

Answer 1

Olá Miqueias,
Tudo bem?
Vamos lá:
$a)sen30^{\circ}=\frac{cat.oposto}{hipotenusa}\\ \\ \frac{1}{2}=\frac{5}{x}\\ \\ x=2\cdot 5\\ \\ \boxed{x=10}\\ \\ b)sen60^{\circ}=\frac{cat.oposto}{hipotenusa}\\ \\ \frac{\sqrt{3}}{2}=\frac{5\sqrt{3}}{x}\\ \\ \sqrt{3}x=2\cdot 5\sqrt{3}\\ \\ x=\frac{10\sqrt{3}}{\sqrt{3}}\\ \\ x=\frac{10\sqrt{3}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ \\ x=\frac{10\sqrt{9}}{\sqrt{9}}\\ \\ x=\frac{10\cdot 3}{3}\\ \\ x=\frac{30}{3}\\ \\ \boxed{x=10}$

$c)sen45^{\circ}=\frac{cat.oposto}{hipotenusa}\\ \\ \frac{\sqrt{2}}{2}=\frac{x}{3\sqrt{2}}\\ \\ 2x=3\sqrt{2\cdot 2}\\ \\ 2x=3\cdot 2\\ \\ 2x=6\\ \\ x=\frac{6}{2}\\ \\ \boxed{x=3}\\ \\ d)tg60^{\circ}=\frac{cat.oposto}{cat.adjacente}\\ \\ \sqrt{3}=\frac{8\sqrt{2}}{x}\\ \\ \sqrt{3}x=8\\ \\ x=\frac{8}{\sqrt{3}}\\ \\ x=\frac{8}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ \\ \boxed{x=\frac{8\sqrt{3}}{3}}$

Espero ter ajudado.