## rebeccaskell94 Group Title Let me draw it. Jim has put a fence along the side AC of the triangular patch of land shown below. 2 years ago 2 years ago

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

what are we trying to find here?

The length of the fence (AC)

use pythogorous theorem no?

5. y2o2 Group Title

(AC)² = (AB)² + (BC)²

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

7. lgbasallote Group Title

what happened there

8. lgbasallote Group Title

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

9. lgbasallote Group Title

dont*

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

11. lgbasallote Group Title

that's not the length of AC though

Do I use Tan?

13. lgbasallote Group Title

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

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

15. lgbasallote Group Title

oops lol....

16. lgbasallote Group Title

sorry that was the distance formula

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

18. lgbasallote Group Title

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

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

20. lgbasallote Group Title

yup