Cos Theta/Sin Theta=10.11; Solve for Theta

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To solve the equation cos(theta)/sin(theta) = 10.11, it can be rewritten as tan(theta) = 1/10.11. Taking the arctangent of both sides provides the value of theta within the interval (0, π/2). The user confirmed the method by using an acot scientific calculator, which yielded consistent results. The discussion highlights a common trigonometric identity and the process of solving for theta. The solution emphasizes the importance of understanding cotangent and tangent relationships in trigonometry.
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if cos(theta)/sin(theta)=10.11, what is theta considering theta is a constant, just spent 2 hours on a collisions question and am stuck on the last hurdle, thanks for looking
 
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Nevermind, ultra blonde moment fml. cotcotcotcotcot!
 
And since cot(θ) = 1/tan(θ), your equation is equivalent to tan(θ) = 1/10.11. Take the arctangent of both sides to find the value of θ in the interval (0, π/2).
 
True that, thanks man, I googled an acot scientific calculator, just checked your method with it and got the same results, boom!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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