Question

Atividade de matemática do Pet 5 do Primeiro ano do Ensino Médio, (Quarta semana).

1- Encontre os valores de x em cada caso.

Answer 1

Explicação passo-a-passo:

a )

$tan60 = \dfrac{8\sqrt{3}}{x}$

$\sqrt{3} = \dfrac{8\sqrt{3}}{x}$

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

$x = \dfrac{8\sqrt{3}}{\sqrt{3}}$

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

b )

Sabendo que 11/2 = 5,5

$sen30 = \dfrac{x}{5,5}$

$\dfrac{1}{2} = \dfrac{x}{5,5}$

$2x = 5,5$

$x = \dfrac{5,5}{2}$

$\boxed{\boxed{ x= 2,75 }}$

c )

$sen45 = \dfrac{6}{x}$

$\dfrac{\sqrt{2}}{2} = \dfrac{6}{x}$

$\sqrt{2} x = 12$

$x = \dfrac{12}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$

$x = \dfrac{12\sqrt{2}}{2}$

$\boxed{\boxed{ x = 6\sqrt{2}}}$

d )

$cos60 = \dfrac{x}{9}$

$\dfrac{1}{2} = \dfrac{x}{9}$

$2x = 9$

$\boxed{\boxed{ x = \dfrac{9}{2}}}$

e )

$cos45 = \dfrac{x}{6}$

$\dfrac{\sqrt{2}}{2} = \dfrac{x}{6}$

$2x = 6\sqrt{2}$

$x = \dfrac{6\sqrt{2}}{2}$

$\boxed{\boxed{ x = 3\sqrt{2}}}$

f )

$cos45 = \dfrac{\sqrt{2}}{y}$

$\dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2}}{y}$

$\sqrt{2} y = 2 \sqrt{2}$

$y = \dfrac{2 \sqrt{2}}{\sqrt{2}}$

$\boxed{\boxed{ y = 2 }}$

$cos30 = \dfrac{2}{x}$

$\dfrac{\sqrt{3}}{2} = \dfrac{2}{x}$

$\sqrt{3} x = 4$

$x = \dfrac{4}{\sqrt{3}}$

$x = \dfrac{4}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

$\boxed{\boxed{ x = \dfrac{4 \sqrt{3}}{3}}}$

Espero ter ajudado !!!