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anonymous

  • one year ago

Find the Value of x and y

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    \[ Side~opposite~30~degrees~=~Short~Leg~=~SL=x\]\[Side~opposite~60~degrees~=~Long~leg~=~LL=y\]\[Hypotenuse~=~H\] \[\large SL~=~\frac{1}{2}H\] \[\large LL~=~SL\sqrt{3}\]

  3. anonymous
    • one year ago
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    im kinda confused tho... how would you solve it

  4. xapproachesinfinity
    • one year ago
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    if x is the opposite what ratio function would you use?

  5. xapproachesinfinity
    • one year ago
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    what do you know about sin and cosine?

  6. anonymous
    • one year ago
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    To solve the short leg (x), plug in H for the first formula.

  7. anonymous
    • one year ago
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    We don't really need sin and cosine for this one, since we have a 30-60-90 triangle. Ez mode

  8. xapproachesinfinity
    • one year ago
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    dear you are using actually that fact!

  9. anonymous
    • one year ago
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    In the diagram \[\large H=4\sqrt{2}\] So plug that into \[\large SL~=~\frac{1}{2}H\]

  10. xapproachesinfinity
    • one year ago
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    where did you get those ratios you are using otherwise

  11. xapproachesinfinity
    • one year ago
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    those numbers you are using came form sin and cos

  12. anonymous
    • one year ago
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    I'm well aware of that

  13. xapproachesinfinity
    • one year ago
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    just wanted to point that since you said no need for sin and cos so just for the poster to be cleared where those came from

  14. anonymous
    • one year ago
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    1/2 * 4 (Sqrt)2

  15. anonymous
    • one year ago
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    Yep. So what would we get from that?

  16. anonymous
    • one year ago
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    2.82

  17. anonymous
    • one year ago
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    Yeah, although it may be easier to leave it in the form of \(2\sqrt{2}\) for now.

  18. anonymous
    • one year ago
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    alright ...

  19. anonymous
    • one year ago
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    Now that we know that \(SL = 2\sqrt{2}\) we can plug it into the next formula \[\large LL~=~SL\sqrt{3}\]

  20. anonymous
    • one year ago
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    so it would be 2sqrt2 * Sqrt 3 ???

  21. anonymous
    • one year ago
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    Yes. \[\large 2\sqrt{2 \times 3}\]

  22. anonymous
    • one year ago
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    oh i got it now !! thank you very very much LegendarySadist !!!

  23. anonymous
    • one year ago
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    \[\huge \color{aqua}N\color{fuchsia}o \space \color{lime}P \color{orange}r \color{blue}o \color{maroon}b \color{red}l \color{olive}e \color{purple}m \ddot\smile \]

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