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March 2, 2015

March 2, 2015

Posted by **Preston** on Sunday, August 28, 2011 at 6:46am.

i have my triangle graphed in the fourth quadrant but I'm getting confused because of the negatives...

- math -
**Reiny**, Sunday, August 28, 2011 at 8:35amcscØ = -3 , so sinØ = -1/3

the sine is negative in III and IV

cosine is positive in I and IV

so you are correct to say that Ø is in IV

make yourself familiar with the CAST rule, here is an explanation for it

http://www.mathsrevision.net/alevel/pages.php?page=36

So your triangle in IV should be a 1-√8 -3 triangle

sinØ = -1/3 , cscØ = -3 (that was the given)

cos = √8/3 , secØ = 3/√8

tanØ = -1/√8 , cotØ = -√8

(Your textbook answer might have written the √8 as 2√2 )

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