I know the radius $R$ of the circle and the area $A$ of the segment.

How can I solve for central angle $\alpha^{\circ}$ in this (or some other) equation:

$$A=\frac{R^{2}}{2} \left( \frac{\alpha \pi}{180}-\sin \alpha^{\circ} \right)$$


Here Newton's algorithm is recommended, but with an initial guess of

$$x(0) = (6k)^{1/3}$$

Why is this the initial guess?

  • $\begingroup$ you after numerical methods I suppose ... $\endgroup$ – Math-fun Mar 30 '16 at 10:13
  • $\begingroup$ @Math-fun Why?? $\endgroup$ – mavavilj Mar 30 '16 at 10:15
  • $\begingroup$ your equation is a mixture of a polynomial and trigonometric function this does not in general admit a closed form solution and one has to rely on numerical methods. $\endgroup$ – Math-fun Mar 30 '16 at 10:17
  • $\begingroup$ I'm looking for if there's some other equation I could use, given this info that I have. Or a pointer into what kind of numerical method to use for the solution? $\endgroup$ – mavavilj Mar 30 '16 at 10:18
  • $\begingroup$ You would get $\alpha -\sin \alpha$= something, use the Taylor approx. for $\sin \alpha$. $\endgroup$ – Nikunj Mar 30 '16 at 10:19

Your equation is transcendental,closed form solution is not possible. Newton -Raphson numerical iteration method is often used. If an approximate solution is acceptable, a graphical solution is also one method.


By series expansion upto 2 terms we get a good approximation

$$ 2 A /R^2 = k \approx \alpha - \sin \alpha = \alpha ^3 /6$$

so we can choose a reasonably accurate value for starting iteration as:

$$ \alpha_{initial}= (6 k)^ { \frac13} .$$

  • $\begingroup$ I really wonder why you received a downvote for a good answer ! $\endgroup$ – Claude Leibovici Mar 31 '16 at 5:58
  • $\begingroup$ @ClaudeLeibovici I initially downvoted the answer before the edit, as I felt the first paragraph, while true, is somewhat redundant to what the OP said or didnt answer what the OP was asking for... The OP didn't want a graphical solution and already knew that Newtons method is used. However, the edit is fantastic, and i have thus removed my down vote and in fact up voted it :) $\endgroup$ – Brevan Ellefsen Mar 31 '16 at 21:15
  • $\begingroup$ Thanks, gentlemen, there was backlash from my side also.. it took time to realize what OP was looking for. $\endgroup$ – Narasimham Mar 31 '16 at 21:22
  • $\begingroup$ @BrevanEllefsen. It is very elegant to explain ! Thanks for acting this way. $\endgroup$ – Claude Leibovici Apr 1 '16 at 4:48

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