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
    24 mins 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
    24 mins 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 5 hours ago









ShibaliciousShibalicious

1295




1295












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


















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
















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




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










1 Answer
1






active

oldest

votes


















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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3124101%2fcalculating-hypotenuses-of-acute-triangles-in-a-circular-segment%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3124101%2fcalculating-hypotenuses-of-acute-triangles-in-a-circular-segment%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness.
            Read more

            Alcedinidae

            Image
            Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
            Read more

            RAC Tourist Trophy

            Image
            Pour les articles homonymes, voir Tourist Trophy. Le RAC Tourist Trophy est une course automobile organisée par le Royal Automobile Club d'Angleterre. Créée en 1905, cette course a fait partie des épreuves comptant pour le championnat du monde des voitures de sport entre 1953 et 1955, en 1958 et 1959, et entre 1962 et 1965. Après une interruption de 17 ans (l'épreuve s'arrête après 1988), le RAC Tourist Trophy reprend en 2005 sous l'égide de la FIA GT. Palmarès | John Napier sur Arrol-Johnston, vainqueur du premier RAC Tourist Trophy, en 1905. L'honorable C.S. Rolls, vainqueur du RAC Tourist Trophy 1906. Ernest Curtis, sur Rover 20 hp, vainqueur du RAC Tourist Trophy 1907. William Watson, vainqueur du RAC Tourist Trophy 1908. Kenelm Lee Guinness, vainqueur du RAC Tourist Trophy 1914 sur Sunbeam. Départ du RAC Tourist Trophy 1928. Louis Gérard, vainqueur du RAC Tourist Trophy 1938 sur Delage. Année Circuit Vainqueur
            Read more