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I am stuck on a problem and can not seem to find a solution, maybe someone here can help me or at least tell me if it is possible to solve?

Please look at the figure:

figure

The problem is: Find the length $x$ given the length $y$ and the arc length $s$. The circle sector angle is unknown but less than 90 degrees.

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closed as off-topic by Travis Willse, Alex Mathers, Shailesh, colormegone, choco_addicted May 19 '16 at 4:04

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  • $\begingroup$ The solution involves $\operatorname{sinc}^{-1}$ which has no simple closed form. $\endgroup$ – kennytm May 18 '16 at 17:02
  • $\begingroup$ I suspected something like that, thank you! $\endgroup$ – Ville May 18 '16 at 17:06
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Suppose the section angle is $\theta$ and radius is $r$, then we have

\begin{align} s &= r\theta, \tag1 \\ y &= r\sin\theta, \tag2 \\ x &= r - r\cos\theta. \tag3 \end{align}

Using (2)/(1) we find that \begin{gather} \frac{\sin\theta}\theta=\frac ys \iff \theta = \operatorname{sinc}^{-1}\frac ys. \\ r =\frac y{\sin\theta} = \frac y{\sin(\operatorname{sinc}^{-1}\frac ys)} \end{gather} Thus $$ x = r(1-\cos\theta) = \frac{y\left(1-\cos\theta\right)}{\sin\theta} = y\tan\frac\theta2=y\tan\left(\frac12\operatorname{sinc}^{-1}\frac ys\right). $$

There is no simple inverse of the sinc function, so we cannot simplify further.

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First let there be an arbitrary radius r.

By the arc length formula, $$r\theta=s$$

Now consider the pythagorean theorem.

$$r^2=y^2+(r-x)^2$$ $$0=y^2-2rx+x^2.$$ $$2rx=y^2+x^2$$ $$r=(y^2+x^2)/2x$$

Therefore $$s=\theta(y^2+x^2)/2x$$ $$2xs/(y^2+x^2)=\theta$$ Now $$sin(2xs/(y^2+x^2))/y=sin(\pi/2)/r$$ $$sin(2xs/(y^2+x^2))=2xy/(x^2+y^2)$$ But $$cos(2xs/(y^2+x^2))=(r-x)/r=1-2x^2/(x^2+y^2)$$ so, by the sin^2+cos^2=1 identity, $$1-4x^2/(x^2+y^2)+4x^4/(x^4+2x^2y^2+y^4)+4x^2y^2/(x^4+2x^2y^2+y^4)=1$$ Simplifying, this is actually just a true statement. The closest we can get to a solution is thus the previous finding: $$sin(2xs/(y^2+x^2))=2xy/(x^2+y^2)$$ This can be used to solve s in terms of x and y, but you need infinite series to solve for x in terms of y and s.

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