Question

determine o valor de seno, cosseno e tangente dos ângulos B e C.

me ajudem, por favor!!!​

Answer 1

a)

${x}^{2} = {6}^{2} + {8}^{2} \\ {x}^{2} = 36 + 64 \\ {x}^{2} = 100 \\ x = \sqrt{100} \\ x = 10$

$\sin(c) = \frac{6}{x} = \frac{6}{10} = \frac{3}{5}$

$\cos(c) = \frac{8}{x} = \frac{8}{10} = \frac{4}{5}$

$\tan(c) = \frac{6}{8} = \frac{3}{4}$

$\sin(b) = \cos(c ) = \frac{4}{5} \\ \cos(b) = \sin(c) = \frac{3}{5} \\ \tan(b) = \frac{1}{ \tan(c) } = \frac{4}{3}$

b)

${x}^{2} = {9}^{2} + {12}^{2} \\ {x}^{2} = 81 + 144 = 225 \\ x = \sqrt{225} \\ x = 15$

$\sin(c) = \frac{9}{x} = \frac{9}{15} = \frac{3}{5}$

$\cos(c) = \frac{12}{x} = \frac{12}{15} = \frac{4}{5}$

$\tan(c) = \frac{9}{12} = \frac{3}{4}$

$\sin(b) = \cos(c) = \frac{4}{5} \\ \cos(b) = \sin(c) = \frac{3}{5} \\ \tan(b) = \frac{1}{ \tan(c) } = \frac{4}{3}$

c) o triângulo é 3,4,5 se tem quatro e 5 é pq falta o 3 portanto

$x = 3$

$\sin(c) = \frac{x}{5} = \frac{3}{5}$

$\cos(c) = \frac{4}{5}$

$\tan(c) = \frac{x}{4} = \frac{3}{4}$

$\sin(b) = \cos(c) = \frac{4}{5} \\ \cos(b) = \sin(c) = \frac{3}{5} \\ \tan(b) = \frac{1}{ \tan(c) } = \frac{4}{3}$

d)

${(2x)}^{2} + {(3x)}^{2} = {20}^{2} \\ 4 {x}^{2} + 9 {x}^{2} = 400 \\ 13 {x}^{2} = 400 \\ {x}^{2} = \frac{400}{13}$

$x = \sqrt{ \frac{400}{13} } = \frac{200}{ \sqrt{13} } = \frac{200 \sqrt{13} }{13}$

$\sin(c) = \frac{2x}{20} = \frac{x}{10} = \frac{1}{10}. \frac{200 \sqrt{13} }{13}$

$\sin(c) = \frac{20 \sqrt{13} }{13}$

$\cos(c) = \frac{3x}{20} = \frac{1}{20} .3x \\ \cos(c) = \frac{1}{20}.3. \frac{200 \sqrt{13} }{13}$

$\cos(c) = \frac{300 \sqrt{13} }{13}$

$\tan(c) = \frac{2x}{3x} = \frac{2}{3}$

$\sin(b) = \cos(c) = \frac{300 \sqrt{13} }{13}$

$\cos(b) = \sin(c) = \frac{20 \sqrt{13} }{13}$

$\tan(b) = \frac{1}{ \tan(c) } = \frac{3}{2}$