I have a sector of a circle split into 16 equal segments. I am trying to calculate the "hypotenuses" of the triangles formed by 2 intersecting straight lines (for example, triangle $LEQ$), which sort encloses the circular sector in rectangular boundaries.

enter image description here


  1. Radius of the circular segment is known.
  2. Angle of the sector (and hence the segments) is known.
  3. Lengths $UE1$ and $JE1$ are known.
  4. $EJ$ is parallel to the X-axis.
  5. Assume α is the angle for each segment.

My approach:

  1. Calculate $KJ$, $KJ = tan(α) \cdot EJ$
  2. From here, line $EK = KJ / sin(α)$
  3. To find the "opposite" of the next segment I do $EJ\cdot(tan(2α) - tan(α))$
  4. Repeat step number 2 with the new value.

The red segments' "opposites" are parallel to the Y-axis. That all works fine until I reach segment 9, where the point $E1$ is. The "opposites" for the blue segments now have to be parallel to the X-axis, and continuing with my approach I only get them parallel with the Y-axis.

Using first calculation as a reference, how can I find the blue "opposites" such as $UT, TS, OE1, NO$, etc..?

EDIT: Great answer! How would one relate length of the radial lines (such as $EN$) in a function of x: f(x) where x = segment number?

  • 1
    $\begingroup$ According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle $\endgroup$ Feb 24 '19 at 2:21
  • $\begingroup$ @J.W.Tanner, I know, my bad, should've put quotation marks around it! $\endgroup$ Feb 25 '19 at 5:58

You then have 16 lines $r_k$, forming angles $k\alpha$ with $x$ axis ($1\le k\le16$).

And then you have two lines $x=x_0$, $y=y_0$, parallel to the axes.

The intersections of lines $r_k$ with these can be readily found as: $$ P_k=(x_0, x_0\tan k\alpha),\quad Q_k=(y_0\cot k\alpha, y_0). $$ In your diagram, $K=P_1$, $R=P_2$, and so on. $A_1$ corresponds to the last $P_k$ whose ordinate is less then $y_0$, that is to: $$ \bar k=\left\lfloor{1\over\alpha}\arctan{y_0\over x_0}\right\rfloor. $$ In the same way, $U=Q_{16}$, $T=Q_{15}$ and so on, down to $\bar k+1$.

To compute $x_0$ and $y_0$ from your data, notice that $y_0=JE_1$ and $UE_1=x_0-y_0\cot 16\alpha$.

  • $\begingroup$ Perfect answer, thank you so much! $\endgroup$ Feb 25 '19 at 6:22
  • $\begingroup$ Sorry, could you please elaborate on why you included $UE1$? Isn't $x0 = JE$ and $y0 = JE1$? $\endgroup$ Feb 25 '19 at 9:24
  • $\begingroup$ What I meant to say was, there are 2 lines parallel to the X-axis, for $x0$ do I use $EJ$ or $UE1$? Or is it $UD1$, if so, why? $\endgroup$ Feb 25 '19 at 9:32
  • $\begingroup$ Also, could you please see my edit? $\endgroup$ Feb 25 '19 at 11:10
  • 1
    $\begingroup$ To find $EN$ use the formula for the distance of two points whose coordinates are known. $\endgroup$ Feb 25 '19 at 18:29

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