• Matéria: Matemática
  • Autor: lelicaki
  • Perguntado 7 anos atrás

urgenteeeee por favorrrrr​

Anexos:

Respostas

respondido por: gibaribi
0

Resposta:

a) cos30°=\frac{10}{x}

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

x\sqrt{3} = 20

\frac{20}{\sqrt{3} } =\frac{\sqrt{3} }{\sqrt{3} }

\frac{20\sqrt{3} }{\sqrt{3} }*\frac{\sqrt{3} }{\sqrt{3} }

x = \frac{20\sqrt{3} }{3}

a^{2}= b^{2}+ c^{2}

(\frac{20\sqrt{3,} }{3}) ^{2} =10^{2} +y^{2}

400=100+y^{2}

y^{2} =300

y=\sqrt{300}

y=10\sqrt{3} }cm

b) cos30°=\frac{x}{20}

\frac{\sqrt{3} }{2} =\frac{x}{20}

2x = 20\sqrt{3

x= \frac{20\sqrt{3} }{2}

x= 10\sqrt{3}cm

a^{2}= b^{2}+ c^{2}

20^{2}= (10\sqrt{3}) ^{2}+ y^{2}

400=300+y^{2}

y^{2} = 400 - 100

y^{2} = 100

y =\sqrt{100}

y=10 cm

c)

cos30° = \frac{x}{40}

\frac{\sqrt{3} }{2}= \frac{x}{40}

2x= 40\sqrt{3}

x= \frac{40\sqrt{3} }{2}

x= 20\sqrt{3} cm

40^{2}= (20\sqrt{3}) ^{2} +y^{2}

1600= 1200 +y^{2}

y^{2} = 400

y= \sqrt{400}

y = 20cm

d)

cos30° = \frac{x}{10}

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

2x = 10\sqrt{3}

x = 5\sqrt{3}  cm

10^{2} =(5\sqrt{3} )^{2} +y^{2}

100 = 75 + y^{2}

y^{2} = 25

y = \sqrt{25}

y = 5

e)

cos30° = \frac{x}{30}

\frac{\sqrt{3} }{2} = \frac{x}{30}

2x = 30 \sqrt{3}

x = \frac{30\sqrt{3} }{2}

x = 15\sqrt{3} cm

30^{2}= (15\sqrt{3} ) ^{2} +y^{2}

900 = 225 + y^{2}

y^{2} = 675

y = \sqrt{675}

y = 15\sqrt{3} cm

f)

cos30° =\frac{8}{x}

\frac{\sqrt{3} }{2} =  \frac{8}{x}

\sqrt{3}*x =16

x= \frac{16}{\sqrt{3} }

x= \frac{16}{\sqrt{16} }= \frac{\sqrt{3} }{\sqrt{3} }

x= \frac{16\sqrt{3} }{3}cm

(\frac{16\sqrt{3} }{y}) ^{2}= 8^{2} +y^{2}

256 = 64 + y^{2}

y^{2} = 192

y = \sqrt{192}

y = 8\sqrt{3} cm

Explicação passo-a-passo:

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