Calculating hypotenuses of acute triangles in a circular segment












2












$begingroup$


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 $EJcdot(tan(2α) - tan(α))$

  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..?






geometry trigonometry triangle circle programming







share|cite|improve this question









$endgroup$












  • $begingroup$
    According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
    $endgroup$
    – J. W. Tanner
    2 hours ago
















2












$begingroup$


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 $EJcdot(tan(2α) - tan(α))$

  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..?






geometry trigonometry triangle circle programming







share|cite|improve this question









$endgroup$












  • $begingroup$
    According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
    $endgroup$
    – J. W. Tanner
    2 hours ago














2












2








2


1



$begingroup$


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 $EJcdot(tan(2α) - tan(α))$

  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..?






geometry trigonometry triangle circle programming







share|cite|improve this question









$endgroup$




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 $EJcdot(tan(2α) - tan(α))$

  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..?






geometry trigonometry triangle circle programming




geometry trigonometry triangle circle programming






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 7 hours ago









ShibaliciousShibalicious

1295




1295












  • $begingroup$
    According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
    $endgroup$
    – J. W. Tanner
    2 hours ago


















  • $begingroup$
    According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
    $endgroup$
    – J. W. Tanner
    2 hours ago
















$begingroup$
According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
$endgroup$
– J. W. Tanner
2 hours ago




$begingroup$
According to Wikipedia, a hypotenuse is the longest side of a right-angled triangle
$endgroup$
– J. W. Tanner
2 hours ago










1 Answer
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$begingroup$

You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).



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:
$$
P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
$$
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:
$$
bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
$$
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
$UE_1=x_0-y_0cot 16alpha$.






share|cite|improve this answer











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    $begingroup$

    You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).



    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:
    $$
    P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
    $$
    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:
    $$
    bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
    $$
    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
    $UE_1=x_0-y_0cot 16alpha$.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).



      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:
      $$
      P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
      $$
      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:
      $$
      bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
      $$
      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
      $UE_1=x_0-y_0cot 16alpha$.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).



        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:
        $$
        P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
        $$
        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:
        $$
        bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
        $$
        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
        $UE_1=x_0-y_0cot 16alpha$.






        share|cite|improve this answer











        $endgroup$



        You then have 16 lines $r_k$, forming angles $kalpha$ with $x$ axis ($1le kle16$).



        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:
        $$
        P_k=(x_0, x_0tan kalpha),quad Q_k=(y_0cot kalpha, y_0).
        $$
        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:
        $$
        bar k=leftlfloor{1overalpha}arctan{y_0over x_0}rightrfloor.
        $$
        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
        $UE_1=x_0-y_0cot 16alpha$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 5 hours ago

























        answered 5 hours ago









        AretinoAretino

        24k21443




        24k21443






























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