Question

Preciso de ajuda para fazer esses cálculos

Answer 1

1)

${8}^{2} = 16n \\ 16n = 64 \\ n = \frac{64}{16} \\ n = 4$

${h}^{2} = 4.12 \\ {h}^{2} = 4.4.3 = 16.3 \\ h = \sqrt{16.3} \\ h = 4 \sqrt{3}$

${c}^{2} = 16.12 \\ {c}^{2} = 16.4.3 \\ c = \sqrt{16.4.3} = 4.2 \sqrt{3} \\ c = 8 \sqrt{3}$

2) seja d a hipotenusa então d=50

${30}^{2} = db \\ 50b = 900 \\ b = \frac{900}{50} \\ b = 18$

$a = d - b = 50 - 18 = 32$

$d.c = 30.40 \\ 50c= 1200 \\ c = \frac{1200}{50} \\ c = 24$

3)

$\sin(30) = \frac{40}{y} \\ \frac{1}{2} = \frac{40}{y} \\ y = 2.40 = 80$

$\tan(45) = \frac{40}{x} \\ 1 = \frac{40}{x} \\ x = 40$

4)

$\tan(30) = \frac{x}{100 + x} \\ \frac{ \sqrt{3} }{3} = \frac{x}{100 + x} \\ 3x = \sqrt{3}(100 + x) \\ 3x = 100 \sqrt{3} +\sqrt{3}x$

$3x - \sqrt{3}x = 100 \sqrt{3} \\ x(3 - \sqrt{3}) = 100 \sqrt{3} \\ x = \frac{100 \sqrt{3} }{3 - \sqrt{3} }$

$x = \frac{100 \sqrt{3}(3 + \sqrt{3}) }{(3 + \sqrt{3})(3 - \sqrt{3})} \\ x = \frac{100 \sqrt{3}(3 + \sqrt{3})}{9 - 3}$

$x = \frac{100 \sqrt{3}(3 + \sqrt{3} }{6} = \frac{50 \sqrt{3}(3 + \sqrt{3})}{3}$

5)

${x}^{2} = {18}^{2} + {10}^{2} - 2.18.10. \cos(60)$

${x}^{2} = 324 + 100 - 2.18.10. \frac{1}{2} \\ {x}^{2} = 424 - 180 \\ {x}^{2} = 244 \\ x = \sqrt{244}$

$x = \sqrt{4.61} \\ x = 2 \sqrt{61}$