Question

Alguém pode me ajudar com isso? ​

Answer 1

Resposta:

Explicação passo-a-passo:

a )

tan60 = \frac{8 \sqrt{3} }{x}tan60=x83

\sqrt{3} = \frac{8 \sqrt{3} } {x}3=x83

\sqrt{3} x = 8 \sqrt{3}3x=83

x = \frac{8 \sqrt{3} }{ \sqrt{3} }x=383

\boxed{\boxed{x = 8}}x=8

b )

sen30 = \frac{x}{ \frac{11}{2} }sen30=211x

\frac{1}{2} = \frac{2x}{11}21=112x

4x = 114x=11

\boxed{\boxed{x = \frac{11}{4}}}x=411

c )

sen45 = \frac{6}{x}sen45=x6

\frac{ \sqrt{2} }{2} = \frac{6}{x}22=x6

\sqrt{2} x = 122x=12

x = \frac{12}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }x=212×22

x = \frac{12 \sqrt{2} }{2} = > \boxed{\boxed{6 \sqrt{2} }}x=2122=>62

d )

cos60 = \frac{x}{9}cos60=9x

\frac{1}{2} = \frac{x}{9}21=9x

2x = 92x=9

\boxed{\boxed{x = \frac{9}{2}}}x=29

e )

cos45 = \frac{x}{6}cos45=6x

\frac{ \sqrt{2} }{2} = \frac{x}{6}22=6x

2x = 6 \sqrt{2}2x