Question

Determine as medidas x e y dos lados dos triângulos abaixo

Answer 1

Explicação passo-a-passo:

Sabendo os valores de seno, cosseno e tangente

a )

$tan45 = \frac{x}{5}$

$1 = \frac{x}{5}$

$x = 5$

Agora descobrimos Y

$cos45 = \frac{5}{y}$

$\frac{ \sqrt{2} }{2} = \frac{5}{y}$

$\sqrt{2} y = 10$

$y = \frac{10}{ \sqrt{2} }$

$y = \frac{10}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }$

$y = \frac{10 \sqrt{2} }{2}$

$y = 5 \sqrt{2}$

b )

$cos30 = \frac{x}{10}$

$\frac{ \sqrt{3} }{2} = \frac{x}{10}$

$2x = 10 \sqrt{3}$

$x = \frac{10 \sqrt{3} }{2}$

$x = 5 \sqrt{3}$

Agora Y

$sen30 = \frac{y}{10}$

$\frac{1}{2} = \frac{y}{10}$

$2y = 10$

$y = 5$

c )

$tan60 = \frac{4}{x}$

$\sqrt{3} = \frac{4}{x}$

$\sqrt{3} x = 4$

$x = \frac{4}{ \sqrt{3} }$

$x = \frac{4}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$x = \frac{4 \sqrt{3} }{3}$

Agora Y

$sen60 = \frac{4}{y}$

$\frac{ \sqrt{3} }{2} = \frac{4}{y}$

$\sqrt{3} y = 8$

$y = \frac{8}{ \sqrt{3} }$

$y = \frac{8}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$y = \frac{ 8 \sqrt{3} }{3}$

Espero ter ajudado !!!