anonymous
  • anonymous
If cos theta = -5/13 and sin theta less than 0, what is tan theta?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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kittiwitti1
  • kittiwitti1
Is this what you're asking?\[\text{If }\cos\theta=-\frac{5}{13}\text{ and }\sin\theta\lt0\text{, what is }\tan\theta?\]
anonymous
  • anonymous
yessss
kittiwitti1
  • kittiwitti1
Okay. First question you ask yourself: which quadrant has negative sines?

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kittiwitti1
  • kittiwitti1
(aka negative "y-values")
anonymous
  • anonymous
okay um 1 and 3 i think
kittiwitti1
  • kittiwitti1
Q1 is wrong, but you got Q3. Here's a reference image: http://www.chilimath.com/algebra/intro/point/images/quadrants_1.gif
anonymous
  • anonymous
i couldnt think without looking so 3 and 4
kittiwitti1
  • kittiwitti1
But good work nonetheless. =) So now you have narrowed down your options to Q3 and Q4.
kittiwitti1
  • kittiwitti1
(sorry I need to find some scratch paper)
Michele_Laino
  • Michele_Laino
hint angle theta has to belongs to the III quadrant: |dw:1435382600490:dw|
Michele_Laino
  • Michele_Laino
belong*
kittiwitti1
  • kittiwitti1
Okay yeah what she said. So now we have a triangle in QIII|dw:1435382756655:dw|
Michele_Laino
  • Michele_Laino
hint: \[\Large \sin \theta = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = ...\]
kittiwitti1
  • kittiwitti1
what
anonymous
  • anonymous
so confusing
kittiwitti1
  • kittiwitti1
lol I don't remember learning that in high school x_x
kittiwitti1
  • kittiwitti1
I thought there were two answers to this question tbh
Michele_Laino
  • Michele_Laino
it is the value of the associated function sin(\theta): \[\Large \sin \theta = - \sqrt {1 - {{\left( {\cos \theta } \right)}^2}} \]
kittiwitti1
  • kittiwitti1
On second thought I think I'll let her explain because I just confused myself. :v
anonymous
  • anonymous
answer choices a. 12/5 b.13/5 c.-12/5 d.square root of 194/5
Michele_Laino
  • Michele_Laino
hint: next steps: \[\Large \begin{gathered} \sin \theta = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = - \sqrt {1 - \frac{{25}}{{169}}} = \hfill \\ \hfill \\ = - \sqrt {\frac{{144}}{{169}}} = - \frac{{12}}{{13}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
still so confused
anonymous
  • anonymous
in case you guys don't get where he got that radical from, he's using the identity \(\sin^2\theta+\cos^2\theta=1\)
Michele_Laino
  • Michele_Laino
hint: so we can write: \[\Large \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{ - \frac{{12}}{{13}}}}{{ - \frac{5}{{13}}}} = ...?\]
kittiwitti1
  • kittiwitti1
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
Michele_Laino
  • Michele_Laino
please, complete my computation above @brandijeannnn
anonymous
  • anonymous
12/5
Michele_Laino
  • Michele_Laino
that's right!

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