SOLUTION: Let θ be an angle in quadrant III, such that csc θ= {{{-9/8}}}
Find tan θ & cos θ:
So far, I found that cosθ = {{{-8/9}}}.
As for tanθ, I think
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Question 985362: Let θ be an angle in quadrant III, such that csc θ=
Find tan θ & cos θ:
So far, I found that cosθ = .
As for tanθ, I think it is
I got that through with x=
And using that, I saw that tanθ's equation would be all over
Right up until I finished typing up this question, I just realized that I misused the reciprocals, oops. I would still like to know the proper answer.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
csc theta = -9/8
note that csc theta = 1 / sin (theta), therefore
1 / sin theta = -9 / 8
sin theta = -8 / 9
inverse sin theta = −62.733955549 approx -62.7
theta is in quadrant III, theta = 270 - 62.7 = 207.3 degrees
*********************************************************************
cos 207.3 = −0.888617233
tan 207.3 = 0.516138488
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