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Mathematics 14 Online
OpenStudy (anonymous):

1.59 = .311/cos^2theta + 11.9sintheta/costheta Solve for theta

OpenStudy (anonymous):

looks better like this \[1.59 = \frac{ .311 }{ \cos ^2\theta } + \frac{ 11.9\sin \theta }{ \cos \theta } \]

OpenStudy (anonymous):

@zepdrix

zepdrix (zepdrix):

Ahh ok I think I got it. Let's see...

zepdrix (zepdrix):

\[\Large\rm 1.59=.311\sec^2\theta+11.9\tan \theta\]\[\Large\rm 1.59=.311(\tan^2\theta+1)+11.9\tan \theta\]And then move some stuff around and you have a quadratic involving tangent.

zepdrix (zepdrix):

\[\Large\rm 0=.311\tan^2\theta+11.9\tan \theta-1.279\]

zepdrix (zepdrix):

\[\Large\rm \tan \theta=\frac{-11.9\pm\sqrt{(11.9)^2-4(.311)(-1.279)}}{2(.311)}\]

OpenStudy (anonymous):

haha I got it! I knew the answer was tan theta = .107 and one the answer given by the quadratic is that while other is negative

zepdrix (zepdrix):

After you get your values on the right, you can apply your inverse tangent to get some angles. Then recall that tangent is periodic in pi, so that gives us a bunch of sets of solutions :d

OpenStudy (anonymous):

this was technically the math portion of a physics problem so all i gotta do is inverse tan .107 and i got my theta :P

zepdrix (zepdrix):

ah neato :)

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