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I am Lyosha [343]
3 years ago
5

Find x on this special right triangle, 45 is not an option!!!!

Mathematics
2 answers:
zloy xaker [14]3 years ago
7 0

let the line between 2 tria be y

sin 60/8√2 = sin 90/y

y=13.06

sin 45/13.06 = sin 90/x

x=18.46

Drupady [299]3 years ago
5 0

Answer:

<u>First, find the hypotenuse of the right triangle with the 60° & 30°.</u>

  • Hypotenuse = h
  • sin(x) = opposite side/hypotenuse

sin(60) = \frac{8\sqrt{2}}{h} \\\\sin(60)h=8\sqrt{2}\\\\\frac{\sqrt{3}}{2} h=8\sqrt{2}\\\\h=\frac{8\sqrt{2}}{\frac{\sqrt{3}}{2}}=8\sqrt{2}*\frac{2}{\sqrt{3}} =\frac{16\sqrt{2} }{\sqrt{3}} =\frac{16\sqrt{2}(\sqrt{3}) }{\sqrt{3}(\sqrt{3})} =\frac{16\sqrt{6} }{3}

<u>Use that side length to find x.</u>

  • sin(x) = opposite side/hypotenuse

sin(45)=\frac{\frac{16\sqrt{6}}{3}}{x}\\\\sin(45)x=\frac{16\sqrt{6}}{3} \\\\\frac{\sqrt{2}}{2}x=\frac{16\sqrt{6}}{3} \\\\x=\frac{\frac{16\sqrt{6}}{3}}{\frac{\sqrt{2}}{2}}=\frac{16\sqrt{6}}{3}*\frac{2}{\sqrt{2}}=\frac{16\sqrt{2}\sqrt{3}(2)}{3\sqrt{2} }=\frac{32\sqrt{3} }{3}

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Answer:

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Step-by-step explanation:

Part a)

step 1

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we have

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substitute

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step 2

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Divide the volume of the sink by the volume of the cup

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Part b)

step 1

Find the volume of the conical cup with a diameter of 8 in. and a height of 8 in

The volume of the cone (cup) is equal to

V=\frac{1}{3}\pi r^{2}h

we have

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assume

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substitute

V=\frac{1}{3}(3.14)(4^{2})8=133.97\ in^3

step 2

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Divide the volume of the sink by the volume of the cup

so

\frac{4,000}{133.97}= 30\ cups

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