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katlin95

  • 3 years ago

please please help:) medals rewarded find x a. 8 b. 8sqrt3 c. 16 sqrt3 d. 1.8 sqrt3

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  1. katlin95
    • 3 years ago
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  2. geoffb
    • 3 years ago
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    You have an angle, as well as hypotenuse and the opposite leg. Which function do you use when you have opposite and hypotenuse: sine, cosine, or tangent?

  3. katlin95
    • 3 years ago
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    this is geometry

  4. geoffb
    • 3 years ago
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    Okay, then you need to know the rules of a 30-60-90 triangle.

  5. geoffb
    • 3 years ago
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    Do you know how the three legs of a triangle relate?

  6. katlin95
    • 3 years ago
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    not reaally but kinda

  7. geoffb
    • 3 years ago
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    The ratio between short leg : long leg : hypotenuse is \(1 : \sqrt{3} : 2\).

  8. geoffb
    • 3 years ago
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    Knowing that, you can use Pythagoras to solve for \(x\). Being across from the 60º angle, would it be the long leg or the short leg?

  9. katlin95
    • 3 years ago
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    long?

  10. geoffb
    • 3 years ago
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    Right.

  11. katlin95
    • 3 years ago
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    so wouldnt the answer be B to this question?

  12. katlin95
    • 3 years ago
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    ?

  13. geoffb
    • 3 years ago
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    That's what I get, yes, since \(\large x = \frac{\sqrt{3}}{2}(h)\)

  14. katlin95
    • 3 years ago
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    wouldnt it be B?

  15. geoffb
    • 3 years ago
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    As I said, yes. \[\large x = \frac{\sqrt{3}}{2}(h) = \frac{16\sqrt{3}}{2} = 8\sqrt{3}\]

  16. katlin95
    • 3 years ago
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    oh sorry their was a freezing problem! thanks so much:)

  17. geoffb
    • 3 years ago
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    No problem. It's being slow on my side, too. Have a good day. :)

  18. katlin95
    • 3 years ago
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    yes you to!!!:)

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