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chris_parks

  • 3 years ago

In a 30-60-90 triangle, the hypotenuse is 20 feet long. Find the length of the length of the long leg and the short leg

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  1. josephniel14
    • 3 years ago
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    short - 10 long - \[10\sqrt{3}\]

  2. satellite73
    • 3 years ago
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    short leg is half of the hypotenuse

  3. josephniel14
    • 3 years ago
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    :))

  4. Sriram
    • 3 years ago
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    20cos60=10 and 20sin60=10 *(30)^1/2

  5. josephniel14
    • 3 years ago
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    @siriam: LOLwut?

  6. josephniel14
    • 3 years ago
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    You can just use some formulas because the given is a special triangle.

  7. satellite73
    • 3 years ago
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    you need to trig for this. short side is half of long side because it is half of an equilateral triangle then the other side comes from pythagoras.

  8. satellite73
    • 3 years ago
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    i meant you need NO trig

  9. Sriram
    • 3 years ago
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    dude 20/short side=cos60 & 20 / long side=sin60 whats wrong in it??

  10. Khu052
    • 3 years ago
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    there's nothing wrong with it, it's just the harder option.

  11. josephniel14
    • 3 years ago
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    Nothing. You can just use simple equations for special triangles.

  12. satellite73
    • 3 years ago
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    nothing wrong except that the only reason you know the cosine of 60 degrees is because you know the ratios, not the other way around!

  13. ARASH_D
    • 3 years ago
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    short=20*sin30=10 long=20*sin60=10\[\sqrt{3}\]

  14. josephniel14
    • 3 years ago
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    Why are you applying trigonometry on such easy questions?

  15. josephniel14
    • 3 years ago
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    You can just use simple equations on special triangles.

  16. josephniel14
    • 3 years ago
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    in a 30-60-90, a = short side, b = long side, c = hypotenuse c = 2a ; b = \[ a \sqrt{3}\]

  17. ARASH_D
    • 3 years ago
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    i think u did it too.i have no problem with trigonometry

  18. robtobey
    • 3 years ago
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    \[\text{long} \text{side}=20\text{Cos}\left[\text{ArcSin}\left[\frac{10}{20}\right]\right]=10 \sqrt{3} \]\[20^2=10^2+\left(10 \sqrt{3}\right)^2, 400=100+300, 400= 400 \]

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