Resposta:
Explicação passo-a-passo:
[tex]sen45°=\frac{co}{h} =\frac{\sqrt{2} }{2} = \frac{y}{20\sqrt{2} }\\2.y=20.\sqrt{4} \\y=\frac{20.2}{2} \\y=20[/tex]
[tex]cos45°=\frac{c.a}{h} =\frac{\sqrt{2} }{2}=\frac{x}{20\sqrt{2} } \\2.x=20.2\\x=20[/tex]
[tex]tang30°=\frac{co}{c.a}=\frac{\sqrt{3} }{3} =\frac{x}{9.\sqrt{3} } \\3.x=9.3\\x=9\\cos30°=\frac{c.a}{h} =\frac{\sqrt{3} }{2} =\frac{9.\sqrt{3} }{y} \\y.\sqrt{3} =18.\sqrt{3} \\y=\frac{18\sqrt{3} }{\sqrt{3} } \\y=18[/tex]