1.59 = .311/cos^2theta + 11.9sintheta/costheta Solve for theta
looks better like this \[1.59 = \frac{ .311 }{ \cos ^2\theta } + \frac{ 11.9\sin \theta }{ \cos \theta } \]
@zepdrix
Ahh ok I think I got it. Let's see...
\[\Large\rm 1.59=.311\sec^2\theta+11.9\tan \theta\]\[\Large\rm 1.59=.311(\tan^2\theta+1)+11.9\tan \theta\]And then move some stuff around and you have a quadratic involving tangent.
\[\Large\rm 0=.311\tan^2\theta+11.9\tan \theta-1.279\]
\[\Large\rm \tan \theta=\frac{-11.9\pm\sqrt{(11.9)^2-4(.311)(-1.279)}}{2(.311)}\]
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
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
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
ah neato :)
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