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anonymous

  • one year ago

If cos theta = -5/13 and sin theta less than 0, what is tan theta?

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  1. kittiwitti1
    • one year ago
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    Is this what you're asking?\[\text{If }\cos\theta=-\frac{5}{13}\text{ and }\sin\theta\lt0\text{, what is }\tan\theta?\]

  2. anonymous
    • one year ago
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    yessss

  3. kittiwitti1
    • one year ago
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    Okay. First question you ask yourself: which quadrant has negative sines?

  4. kittiwitti1
    • one year ago
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    (aka negative "y-values")

  5. anonymous
    • one year ago
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    okay um 1 and 3 i think

  6. kittiwitti1
    • one year ago
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    Q1 is wrong, but you got Q3. Here's a reference image: http://www.chilimath.com/algebra/intro/point/images/quadrants_1.gif

  7. anonymous
    • one year ago
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    i couldnt think without looking so 3 and 4

  8. kittiwitti1
    • one year ago
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    But good work nonetheless. =) So now you have narrowed down your options to Q3 and Q4.

  9. kittiwitti1
    • one year ago
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    (sorry I need to find some scratch paper)

  10. Michele_Laino
    • one year ago
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    hint angle theta has to belongs to the III quadrant: |dw:1435382600490:dw|

  11. Michele_Laino
    • one year ago
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    belong*

  12. kittiwitti1
    • one year ago
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    Okay yeah what she said. So now we have a triangle in QIII|dw:1435382756655:dw|

  13. Michele_Laino
    • one year ago
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    hint: \[\Large \sin \theta = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = ...\]

  14. kittiwitti1
    • one year ago
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    what

  15. anonymous
    • one year ago
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    so confusing

  16. kittiwitti1
    • one year ago
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    lol I don't remember learning that in high school x_x

  17. kittiwitti1
    • one year ago
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    I thought there were two answers to this question tbh

  18. Michele_Laino
    • one year ago
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    it is the value of the associated function sin(\theta): \[\Large \sin \theta = - \sqrt {1 - {{\left( {\cos \theta } \right)}^2}} \]

  19. kittiwitti1
    • one year ago
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    On second thought I think I'll let her explain because I just confused myself. :v

  20. anonymous
    • one year ago
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    answer choices a. 12/5 b.13/5 c.-12/5 d.square root of 194/5

  21. Michele_Laino
    • one year ago
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    hint: next steps: \[\Large \begin{gathered} \sin \theta = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = - \sqrt {1 - \frac{{25}}{{169}}} = \hfill \\ \hfill \\ = - \sqrt {\frac{{144}}{{169}}} = - \frac{{12}}{{13}} \hfill \\ \end{gathered} \]

  22. anonymous
    • one year ago
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    still so confused

  23. anonymous
    • one year ago
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    in case you guys don't get where he got that radical from, he's using the identity \(\sin^2\theta+\cos^2\theta=1\)

  24. Michele_Laino
    • one year ago
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    hint: so we can write: \[\Large \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{ - \frac{{12}}{{13}}}}{{ - \frac{5}{{13}}}} = ...?\]

  25. kittiwitti1
    • one year ago
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    Oh I mostly get it but I was multitasking sorry xD so I decided to leave the problem to someone who could focus their whole effort into helping

  26. Michele_Laino
    • one year ago
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    please, complete my computation above @brandijeannnn

  27. anonymous
    • one year ago
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    12/5

  28. Michele_Laino
    • one year ago
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    that's right!

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