SOLUTION: If tanx= 5/3 and secx < 0, then what is the exact value of sinx? So far, I have: Opp: 5, Adj; 3, and Hyp: sqrt(34). Sin^2x= sinxcosx But after that I'm stuck! P

Algebra ->  Trigonometry-basics -> SOLUTION: If tanx= 5/3 and secx < 0, then what is the exact value of sinx? So far, I have: Opp: 5, Adj; 3, and Hyp: sqrt(34). Sin^2x= sinxcosx But after that I'm stuck! P      Log On


   



Question 716473: If tanx= 5/3 and secx < 0, then what is the exact value of sinx?
So far, I have:

Opp: 5, Adj; 3, and Hyp: sqrt(34). Sin^2x= sinxcosx
But after that I'm stuck! Please help!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If tanx= 5/3 and secx < 0, then what is the exact value of sinx?
So far, I have:
Opp: 5, Adj; 3, and Hyp: sqrt(34). Sin^2x= sinxcosx
But after that I'm stuck! Please help!
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Since tan is positive and sec is negative x in an angle in QIII
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tan = y/x = -5/-3, so y = -5, x = -3
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Solve for "r": r = sqrt[5^2+3^2] = sqrt(34)
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Ans: sin(x) = y/r = -5/sqrt(34)
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Cheers,
Stan H.
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