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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

Conditions:

  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?

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    $\begingroup$ According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle $\endgroup$ – J. W. Tanner Feb 24 '19 at 2:21
  • $\begingroup$ @J.W.Tanner, I know, my bad, should've put quotation marks around it! $\endgroup$ – Shibalicious Feb 25 '19 at 5:58
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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 $UE_1=x_0-y_0\cot 16\alpha$.

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  • $\begingroup$ Perfect answer, thank you so much! $\endgroup$ – Shibalicious Feb 25 '19 at 6:22
  • $\begingroup$ Sorry, could you please elaborate on why you included $UE1$? Isn't $x0 = JE$ and $y0 = JE1$? $\endgroup$ – Shibalicious 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? $\endgroup$ – Shibalicious Feb 25 '19 at 9:32
  • $\begingroup$ Also, could you please see my edit? $\endgroup$ – Shibalicious Feb 25 '19 at 11:10
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    $\begingroup$ To find $EN$ use the formula for the distance of two points whose coordinates are known. $\endgroup$ – Aretino Feb 25 '19 at 18:29

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