Apologies for what is probably an easy question but my mind has gone black, I need to rearrange this equation for $b$: \begin{equation} a=\left(\frac{180}{\pi}\right)\arccos \left(\frac{b-cd}{b}\right) \end{equation}
Thanks in advance
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$\begingroup$ damn, clicked on the wrong tag! sorry! what would it go under? $\endgroup$– bolt19Nov 13, 2014 at 15:17
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$$\begin{align*} a &= \frac{180}{\pi}\arccos\frac{b-cd}{b}\\ \arccos\left(1-\frac{cd}{b}\right) &= \frac{a\pi}{180}\\ 1-\frac{cd}{b} &= \cos\frac{a\pi}{180}\\ \frac{cd}b &= 1-\cos\frac{a\pi}{180}\\ b &= \frac{cd}{1-\cos\frac{a\pi}{180}} \end{align*}$$
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$\begingroup$ Was it originally an $\arccos$ in the inital question or $\cos$ multiplied by a constant $A$? $\endgroup$ Nov 13, 2014 at 15:28
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$\begingroup$ I also would like @bolt19 to confirm. The above was based on how WolframAlpha interpreted
a = Acos((b - (c * d)) / b) * 180 / PI$\endgroup$– peterwhyNov 13, 2014 at 15:31 -
$\begingroup$ it was arccos in the initial question, and thanks, I will use it in the future! $\endgroup$– bolt19Nov 13, 2014 at 15:32
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