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lala2

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  • 10 months ago
  • 10 months ago

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  1. GoldPhenoix
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    Hello. Do you need help?

    • 10 months ago
  2. lala2
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    use a unit circle and 30 60 90 triangle to find value in degrees.. tan ^-1 sqrt 3

    • 10 months ago
  3. lala2
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    |dw:1369790876325:dw|

    • 10 months ago
  4. lala2
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    can someone help

    • 10 months ago
  5. lala2
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    @onedirection822

    • 10 months ago
  6. lala2
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    @arykk

    • 10 months ago
  7. satellite73
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    look on the unit circle to find where the second coordinate is \(-\frac{\sqrt{3}}{2}\) and the first coordinate is \(\frac{1}{2}\)

    • 10 months ago
  8. satellite73
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    oops sorry, should be second coordinate is \(\frac{\sqrt{3}}{2}\) and first coordinate is \(\frac{1}{2}\)

    • 10 months ago
  9. lala2
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    what does that look like?

    • 10 months ago
  10. satellite73
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    you will find it on the last page of the attached cheat sheet

    • 10 months ago
  11. lala2
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    |dw:1369791474788:dw|if you have a unit circle like this

    • 10 months ago
  12. satellite73
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    unit blob? look at the cheat sheet i sent, on the last page you see the unit circle

    • 10 months ago
  13. lala2
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    haah! thats mean!

    • 10 months ago
  14. lala2
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    ok ... but where exaclty is it?

    • 10 months ago
  15. satellite73
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    look at the angle of \(\frac{\pi}{3}\) and you will see that the second coordinate is \(\frac{\sqrt{3}}{2}\) and the first coordinate is \(\frac{1}{2}\) and so \[\tan(\frac{\pi}{3})=\sqrt3\]

    • 10 months ago
  16. lala2
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    i dont understand all of the symbols

    • 10 months ago
  17. lala2
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    @satellite73 right here

    • 10 months ago
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  18. lala2
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    @satellite73 what do i do next

    • 10 months ago
  19. satellite73
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    that is all

    • 10 months ago
  20. lala2
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    but it says to use 30 60 90 triangle

    • 10 months ago
  21. satellite73
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    oh you are working in degrees. not problem since \(\tan(60)=\sqrt3\) you know \(\tan^{-1}(\sqrt3)=60\)

    • 10 months ago
  22. satellite73
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    |dw:1369791928159:dw|

    • 10 months ago
  23. satellite73
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    tangent is opposite over adjacent is \(\sqrt3\)

    • 10 months ago
  24. lala2
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    what about cos^-1 (sqrt3)/2

    • 10 months ago
  25. lala2
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    @satellite73

    • 10 months ago
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