Question

1- Encontre os valores de x em cada caso.​

Answer 1

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a)

$\sf tg(60^\circ)=\dfrac{8\sqrt{3}}{x}\\\sf\diagup\!\!\!\!\!\!\sqrt{3}=\dfrac{8\diagup\!\!\!\!\!\!\sqrt{3}}{x}\\\sf x=8$

b)

$\sf sen(30^\circ)=\dfrac{x}{\frac{11}{2}}\\\sf x=\dfrac{11}{2}\cdot sen(30^\circ)\\\sf x=\dfrac{11}{2}\cdot\dfrac{1}{2}\\\sf x=\dfrac{11}{4}$

c)

$\sf sen(45^\circ)=\dfrac{6}{x}\\\sf\dfrac{\sqrt{2}}{2}=\dfrac{6}{x}\\\sf\sqrt{2}\cdot x=6\cdot2\\\sf x=\dfrac{12}{\sqrt{2}}\\\sf x=\dfrac{12\sqrt{2}}{2}\\\sf x=6\sqrt{2}$

d)

$\sf cos(60^\circ)=\dfrac{x}{9}\\\sf\dfrac{1}{2}=\dfrac{x}{9}\\\sf 2x=9\\\sf x=\dfrac{9}{2}$

e)

 $\sf cos(45^\circ)=\dfrac{x}{6}\\\sf\dfrac{\sqrt{2}}{2}=\dfrac{x}{6}\\\sf2x=6\sqrt{2}\\\sf x=\dfrac{6\sqrt{2}}{2}\\\sf x=3\sqrt{2}$

f) observe a figura que anexei.

vamos usar o teorema de Pitágoras para calcular o valor de y:

$\sf y^2=(\sqrt{2})^2+(\sqrt{10})^2\\\sf y^2=2+10\\\sf y^2=12\\\sf y^2=4\cdot3\\\sf y=\sqrt{4\cdot3}\\\sf y=2\sqrt{3}$

Agora vamos a relação cosseno para calcular x e finalizar a questão :$\sf cos(30^\circ)=\dfrac{y}{x}\\\sf\dfrac{\diagup\!\!\!\!\!\!\sqrt{3}}{2}=\dfrac{2\diagup\!\!\!\!\!\!\sqrt{3}}{x}\\\sf x=2\cdot2\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=2\checkmark}}}}$