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rebeccaskell94

Let me draw it. Jim has put a fence along the side AC of the triangular patch of land shown below.

  • one year ago
  • one year ago

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  1. rebeccaskell94
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    |dw:1342705835156:dw| To solve this do I do tan(x) = 8 ÷ 15? and if not what should I do?

    • one year ago
  2. vamgadu
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    what are we trying to find here?

    • one year ago
  3. rebeccaskell94
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    The length of the fence (AC)

    • one year ago
  4. vamgadu
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    use pythogorous theorem no?

    • one year ago
  5. y2o2
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    (AC)² = (AB)² + (BC)²

    • one year ago
  6. rebeccaskell94
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    oh just 8^2 + 15^2 = 64 + 225 = 289 or √17

    • one year ago
  7. lgbasallote
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    what happened there

    • one year ago
  8. lgbasallote
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    sont you mean 64 + 225 = 289 or 17^2

    • one year ago
  9. lgbasallote
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    dont*

    • one year ago
  10. rebeccaskell94
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    Oh yeah sory I meant 17^2 Is that no bueno?

    • one year ago
  11. lgbasallote
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    that's not the length of AC though

    • one year ago
  12. rebeccaskell94
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    Do I use Tan?

    • one year ago
  13. lgbasallote
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    \[AC = \sqrt{x^2 + y^2}\]

    • one year ago
  14. rebeccaskell94
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    I don't know what x and y are, though? Unless 8 and 15 are x and y?

    • one year ago
  15. lgbasallote
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    oops lol....

    • one year ago
  16. lgbasallote
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    sorry that was the distance formula

    • one year ago
  17. rebeccaskell94
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    haha no worries. I'm just glad you're helping :D

    • one year ago
  18. lgbasallote
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    \[AC = \sqrt{AB^2 + BC^2}\]

    • one year ago
  19. rebeccaskell94
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    Ah! Okay! So, 225+64 = √289? so the length of AC is 17?

    • one year ago
  20. lgbasallote
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    yup

    • one year ago
  21. rebeccaskell94
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    Thank you!

    • one year ago
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