Question

Me ajudem só no 1 a e b

Answer 1

Explicação passo-a-passo:

a)

$sen45 = \frac{co}{h} \\ \frac{ \sqrt{2} }{2} = \frac{y}{20 \sqrt{2} } \\ 2y = 20 \sqrt{4} \\ 2y = 20 \times 2 \\ 2y = 40 \\ y = \frac{40}{2} \\ y = 20$

$cos45 = \frac{ca}{h} \\ \frac{ \sqrt{2} }{2} = \frac{x}{20 \sqrt{2} } \\ 2x = 20 \sqrt{4} \\ 2x = 20 \times 2 \\ 2x = 40 \\ x = \frac{40}{2} \\ x = 20$

b)

$tg30 = \frac{co}{ca} \\ \frac{ \sqrt{3} }{3} = \frac{x}{9 \sqrt{3} } \\ 3x = 9 \sqrt{9} \\ 3x = 9 \times 3 \\ x = \frac{9 \times 3}{3} \\ x = 9$

$cos30 = \frac{ca}{h} \\ \frac{ \sqrt{3} }{2} = \frac{9 \sqrt{3} }{y} \\ \sqrt{3} y = 18 \sqrt{3} \\ y = \frac{18 \sqrt{3} }{ \sqrt{3} } \\ y = 18$