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Solve for θ.
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For which problem?
crap hang on.
\[mg \sin \theta = \mu mg \cos \theta ((M+m)/(m))\]
divide both sides by cos(theta) and divide both sides by mg
tan(theta)mg=um(M+m)/m tan(theta)=u(M+m)/m theta=tan^-1(u(M+m)/m) someone check me
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then you get \[\frac{\sin(\theta)}{\cos(\theta)}=\mu \frac{\not{m}\not{g}}{\not{m}\not{g}} \frac{M+m}{m}\]
\[\tan(\theta)=\mu \frac{M+m}{m} => \theta=\tan^{-1}(\mu \frac{M+m}{m})\]
whoa where di tan theta come from?!?!
*did
divided both sides by cos(theta) because remeber sin/cos=tan from trig?
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oh ok its been awhile lol
i know we all have that moment, haha
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