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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

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

2. vamgadu

what are we trying to find here?

3. rebeccaskell94

The length of the fence (AC)

4. vamgadu

use pythogorous theorem no?

5. y2o2

(AC)² = (AB)² + (BC)²

6. rebeccaskell94

oh just 8^2 + 15^2 = 64 + 225 = 289 or √17

7. lgbasallote

what happened there

8. lgbasallote

sont you mean 64 + 225 = 289 or 17^2

9. lgbasallote

dont*

10. rebeccaskell94

Oh yeah sory I meant 17^2 Is that no bueno?

11. lgbasallote

that's not the length of AC though

12. rebeccaskell94

Do I use Tan?

13. lgbasallote

$AC = \sqrt{x^2 + y^2}$

14. rebeccaskell94

I don't know what x and y are, though? Unless 8 and 15 are x and y?

15. lgbasallote

oops lol....

16. lgbasallote

sorry that was the distance formula

17. rebeccaskell94

haha no worries. I'm just glad you're helping :D

18. lgbasallote

$AC = \sqrt{AB^2 + BC^2}$

19. rebeccaskell94

Ah! Okay! So, 225+64 = √289? so the length of AC is 17?

20. lgbasallote

yup

21. rebeccaskell94

Thank you!

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