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rebeccaskell94

  • 2 years ago

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

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

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

  3. rebeccaskell94
    • 2 years ago
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    The length of the fence (AC)

  4. vamgadu
    • 2 years ago
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    use pythogorous theorem no?

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

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

  7. lgbasallote
    • 2 years ago
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    what happened there

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

  9. lgbasallote
    • 2 years ago
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    dont*

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

  11. lgbasallote
    • 2 years ago
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    that's not the length of AC though

  12. rebeccaskell94
    • 2 years ago
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    Do I use Tan?

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

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

  15. lgbasallote
    • 2 years ago
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    oops lol....

  16. lgbasallote
    • 2 years ago
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    sorry that was the distance formula

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

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

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

  20. lgbasallote
    • 2 years ago
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    yup

  21. rebeccaskell94
    • 2 years ago
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    Thank you!

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