Question

Me ajudem, pra hoje, obrigada!!!

Answer 1

Como o triângulo retângulo ABD é isósceles, temos $AD=AB=2$. Pelo Teorema de Pitágoras,

$BD^2=2^2+2^2~~\Rightarrow~~BD^2=4+4~~\Rightarrow~~BD^2=8$.

$BD=\sqrt{8}~~\Rightarrow~~\BD=2\sqrt{2}$.

Temos que:

$\text{sen}~30^{\circ}=\dfrac{x}{2\sqrt{2}}$

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

$2x=2\sqrt{2}$

$x=\sqrt{2}$


$\text{cos}~30^{\circ}=\dfrac{y}{2\sqrt{2}}$

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

$2y=2\sqrt{6}$

$y=\sqrt{6}$.

Answer 2

$BD = 2 \sqrt{2} \\ \\ sen30=\frac{x}{2\sqrt2}\\ \\ \frac{1}{2}=\frac{x}{2\sqrt2}\\ \\ x=\sqrt2\\ \\ y^2=(2\sqrt2)^2-(\sqrt2)^2\\ \\ y^2=4-2\\ \\ y=\sqrt2$