# Calculating "hypotenuses" of acute triangles in a circular segment

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.

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?

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

• Perfect answer, thank you so much! Feb 25 '19 at 6:22
• Sorry, could you please elaborate on why you included $UE1$? Isn't $x0 = JE$ and $y0 = JE1$? Feb 25 '19 at 9:24
• 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? Feb 25 '19 at 9:32
• Also, could you please see my edit? Feb 25 '19 at 11:10
• To find $EN$ use the formula for the distance of two points whose coordinates are known. Feb 25 '19 at 18:29