Question

calcule o valor de x nos triângulos retângulos (Matéria Trigonometria),
me ajudem pfv!!!!

Answer 1

Oi

a)

sen30° = cat.op / hip

1 / 2 = 2 / x 

x = 2*2

x = 4 

b)

sen45° = cat.op / hip

√2 / 2 = x / 4 

2x = 4√2

x = 4√2 / 2 

x = 2√2

c)

cos15° = cat.aj / hip

0,96 = 2 / x 

x = 2 / 0,96

x = 2,08

Answer 2

a)

$\color{red}{\sin30\degree}=\color{blue}{\frac{2}{x}}$

$\color{blue}{\frac{1}{2}}=\color{blue}{\frac{2}{x}}$

$\color{blue}{x=2.2=4}\color{green}{\checkmark}$

b)

$\color{red}{\sin45\degree}=\color{blue}{\frac{x}{4}}$

$\color{blue}{\frac{\sqrt{2}}{2}}=\color{blue}{\frac{x}{4}}$

$\color{blue}{2x=4\sqrt{2}}\\\color{blue}{x=\frac{4\sqrt{2}}{2}}\\\color{blue}{x=2\sqrt{2}}\color{green}{\checkmark}$

c)

$\cos(15\degree)=\cos(45\degree-30\degree)\\=\cos(45\degree).\cos(30\degree)+\sin(45\degree).\sin(30\degree)$

$\cos(15\degree)=\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}.\frac{1}{2}$

$\cos(15\degree)=\frac{\sqrt{6}+\sqrt{2}}{4}$

$\color{red}{\cos(15\degree)}=\color{dodgerblue}{\frac{2}{x}}$

$\color{red}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\color{dodgerblue}{\frac{2}{x}}$

$\color{dodgerblue}{(\sqrt{6}+\sqrt{2}).x=4.2}$

$\color{dodgerblue}{x=\frac{8}{(\sqrt{6}+\sqrt{2})}}$

$\color{dodgerblue}{x=\frac{8}{(\sqrt{6}+\sqrt{2})}.\frac{(\sqrt{6}-\sqrt{2})}{(\sqrt{6}-\sqrt{2})}}$

$\color{dodgerblue}{x=\frac{8(\sqrt{6}-\sqrt{2})}{{(\sqrt{6})}^{2}-{(\sqrt{2})}^{2}}}$

$\color{dodgerblue}{\frac{8(\sqrt{6}+\sqrt{2})}{6-2}}$

$\color{dodgerblue}{\frac{\cancel{8}(\sqrt{6}+\sqrt{2})}{\cancel{4}}}$

$\color{dodgerblue}{x=2(\sqrt{6}+\sqrt{2})}\color{green}{\checkmark}$