Calculating hypotenuses of acute triangles in a circular segment












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









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    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









    $endgroup$















      1












      1








      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











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      asked 4 hours ago









      ShibaliciousShibalicious

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      1245






















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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











          $endgroup$













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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 3 hours ago

























                answered 3 hours ago









                AretinoAretino

                24k21443




                24k21443






























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