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I have a literal equation that needs to be solved for $\theta$:

$$mg \sin(\theta) = \mu mg \cos(\theta)\left({ M+m \over m}\right) $$

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Do you mean $mg \sin \theta = umg \cos \theta \cdot \frac{M+m}{m}$? – nayrb Aug 28 '12 at 0:04
Use $\sin x/\cos x=\tan x$ and the arctangent function. – Gerry Myerson Aug 28 '12 at 0:05
@jay: Did I get it right? If so, you can use the "edit" button to see what I did to change the formatting. If not, say what's wrong and someone will fix it. – MJD Aug 28 '12 at 0:06
What is a «literal equation»? – Mariano Suárez-Alvarez Aug 28 '12 at 0:08
@MarianoSuárez-Alvarez It's an equation whose coefficients are letters instead of numbers, I guess. Such as $ax+b=0$, but not $2x-5=0$. – Américo Tavares Aug 28 '12 at 0:13
up vote 6 down vote accepted

A simple rearrangement shows that $$\tan\theta = \mu\left(1+\frac{M}{m}\right)$$ to obtain numerical values for the equation you'll need the ratio of the masses and the coefficient of friction, in which case you can simply take the arctangent.

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