anonymous
  • anonymous
Let me draw it. Jim has put a fence along the side AC of the triangular patch of land shown below.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1342705835156:dw| To solve this do I do tan(x) = 8 ÷ 15? and if not what should I do?
anonymous
  • anonymous
what are we trying to find here?
anonymous
  • anonymous
The length of the fence (AC)

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anonymous
  • anonymous
use pythogorous theorem no?
y2o2
  • y2o2
(AC)² = (AB)² + (BC)²
anonymous
  • anonymous
oh just 8^2 + 15^2 = 64 + 225 = 289 or √17
lgbasallote
  • lgbasallote
what happened there
lgbasallote
  • lgbasallote
sont you mean 64 + 225 = 289 or 17^2
lgbasallote
  • lgbasallote
dont*
anonymous
  • anonymous
Oh yeah sory I meant 17^2 Is that no bueno?
lgbasallote
  • lgbasallote
that's not the length of AC though
anonymous
  • anonymous
Do I use Tan?
lgbasallote
  • lgbasallote
\[AC = \sqrt{x^2 + y^2}\]
anonymous
  • anonymous
I don't know what x and y are, though? Unless 8 and 15 are x and y?
lgbasallote
  • lgbasallote
oops lol....
lgbasallote
  • lgbasallote
sorry that was the distance formula
anonymous
  • anonymous
haha no worries. I'm just glad you're helping :D
lgbasallote
  • lgbasallote
\[AC = \sqrt{AB^2 + BC^2}\]
anonymous
  • anonymous
Ah! Okay! So, 225+64 = √289? so the length of AC is 17?
lgbasallote
  • lgbasallote
yup
anonymous
  • anonymous
Thank you!

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