How to use double angle identities to find length 'd'?












0












$begingroup$


Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.



A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
See illustration.



Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.



By using $$cos 2x=2cos^2 x-1$$
My working is:
$$cos x=frac d3$$
$$cos 2x=frac d3$$
therefore:
$$frac d3=2(frac d3)^2-1$$



Now becomes solvable as a quadratic equation






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
    $endgroup$
    – hardmath
    Dec 11 '18 at 16:39










  • $begingroup$
    Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
    $endgroup$
    – HawkinsCivil
    Dec 11 '18 at 16:40












  • $begingroup$
    This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
    $endgroup$
    – Love Invariants
    Dec 11 '18 at 16:53










  • $begingroup$
    Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
    $endgroup$
    – HawkinsCivil
    Dec 11 '18 at 16:56










  • $begingroup$
    Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
    $endgroup$
    – Love Invariants
    Dec 11 '18 at 17:24


















0












$begingroup$


Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.



A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
See illustration.



Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.



By using $$cos 2x=2cos^2 x-1$$
My working is:
$$cos x=frac d3$$
$$cos 2x=frac d3$$
therefore:
$$frac d3=2(frac d3)^2-1$$



Now becomes solvable as a quadratic equation






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
    $endgroup$
    – hardmath
    Dec 11 '18 at 16:39










  • $begingroup$
    Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
    $endgroup$
    – HawkinsCivil
    Dec 11 '18 at 16:40












  • $begingroup$
    This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
    $endgroup$
    – Love Invariants
    Dec 11 '18 at 16:53










  • $begingroup$
    Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
    $endgroup$
    – HawkinsCivil
    Dec 11 '18 at 16:56










  • $begingroup$
    Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
    $endgroup$
    – Love Invariants
    Dec 11 '18 at 17:24
















0












0








0


1



$begingroup$


Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.



A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
See illustration.



Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.



By using $$cos 2x=2cos^2 x-1$$
My working is:
$$cos x=frac d3$$
$$cos 2x=frac d3$$
therefore:
$$frac d3=2(frac d3)^2-1$$



Now becomes solvable as a quadratic equation






trigonometry







share|cite|improve this question











$endgroup$




Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.



A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
See illustration.



Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.



By using $$cos 2x=2cos^2 x-1$$
My working is:
$$cos x=frac d3$$
$$cos 2x=frac d3$$
therefore:
$$frac d3=2(frac d3)^2-1$$



Now becomes solvable as a quadratic equation






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 22:01







HawkinsCivil

















asked Dec 11 '18 at 16:08









HawkinsCivilHawkinsCivil

14




14












  • $begingroup$
    I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
    $endgroup$
    – hardmath
    Dec 11 '18 at 16:39










  • $begingroup$
    Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
    $endgroup$
    – HawkinsCivil
    Dec 11 '18 at 16:40












  • $begingroup$
    This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
    $endgroup$
    – Love Invariants
    Dec 11 '18 at 16:53










  • $begingroup$
    Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
    $endgroup$
    – HawkinsCivil
    Dec 11 '18 at 16:56










  • $begingroup$
    Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
    $endgroup$
    – Love Invariants
    Dec 11 '18 at 17:24




















  • $begingroup$
    I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
    $endgroup$
    – hardmath
    Dec 11 '18 at 16:39










  • $begingroup$
    Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
    $endgroup$
    – HawkinsCivil
    Dec 11 '18 at 16:40












  • $begingroup$
    This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
    $endgroup$
    – Love Invariants
    Dec 11 '18 at 16:53










  • $begingroup$
    Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
    $endgroup$
    – HawkinsCivil
    Dec 11 '18 at 16:56










  • $begingroup$
    Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
    $endgroup$
    – Love Invariants
    Dec 11 '18 at 17:24


















$begingroup$
I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
$endgroup$
– hardmath
Dec 11 '18 at 16:39




$begingroup$
I improved the math formatting a little. You can read more about using MathJax and $LaTeX$ in this introduction.
$endgroup$
– hardmath
Dec 11 '18 at 16:39












$begingroup$
Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:40






$begingroup$
Yes I noticed, thanks. This is my first post so still getting my head around the formatting side of it. How did you get this image to display within the post?
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:40














$begingroup$
This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
$endgroup$
– Love Invariants
Dec 11 '18 at 16:53




$begingroup$
This question needs explanation especially for non-english readers. I get nothing from it. Please explain it more if you can.
$endgroup$
– Love Invariants
Dec 11 '18 at 16:53












$begingroup$
Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:56




$begingroup$
Which part needs further explanation? I guess part of the problem is that I don't fully understand it myself or know if my approach is correct, sorry
$endgroup$
– HawkinsCivil
Dec 11 '18 at 16:56












$begingroup$
Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
$endgroup$
– Love Invariants
Dec 11 '18 at 17:24






$begingroup$
Ok, I got it. I was unable to locate the pole. Now located. Tell me one thing, Is A not in air? I mean if pole is vertical fixed on ground.
$endgroup$
– Love Invariants
Dec 11 '18 at 17:24












3 Answers
3






active

oldest

votes


















0












$begingroup$

I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"



Using the law of cosines in triangles $ABC$ and $ACD$ we get
$$c^2=2^2+3^2-2times 2times 3 cos x$$
$$b^2=d^2+2^2-4dcos x$$
so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
The last equation is quadratic in $d$ and easy to solve.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
    $endgroup$
    – BPP
    Dec 11 '18 at 20:09



















0












$begingroup$

To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
$$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    $cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.






    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%2f3035443%2fhow-to-use-double-angle-identities-to-find-length-d%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"



      Using the law of cosines in triangles $ABC$ and $ACD$ we get
      $$c^2=2^2+3^2-2times 2times 3 cos x$$
      $$b^2=d^2+2^2-4dcos x$$
      so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
      The last equation is quadratic in $d$ and easy to solve.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
        $endgroup$
        – BPP
        Dec 11 '18 at 20:09
















      0












      $begingroup$

      I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"



      Using the law of cosines in triangles $ABC$ and $ACD$ we get
      $$c^2=2^2+3^2-2times 2times 3 cos x$$
      $$b^2=d^2+2^2-4dcos x$$
      so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
      The last equation is quadratic in $d$ and easy to solve.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
        $endgroup$
        – BPP
        Dec 11 '18 at 20:09














      0












      0








      0





      $begingroup$

      I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"



      Using the law of cosines in triangles $ABC$ and $ACD$ we get
      $$c^2=2^2+3^2-2times 2times 3 cos x$$
      $$b^2=d^2+2^2-4dcos x$$
      so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
      The last equation is quadratic in $d$ and easy to solve.






      share|cite|improve this answer









      $endgroup$



      I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"



      Using the law of cosines in triangles $ABC$ and $ACD$ we get
      $$c^2=2^2+3^2-2times 2times 3 cos x$$
      $$b^2=d^2+2^2-4dcos x$$
      so that $$cos x=dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-dfrac{4d(13-c^2)}{12}$$
      The last equation is quadratic in $d$ and easy to solve.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 11 '18 at 17:29









      BPPBPP

      2,169927




      2,169927












      • $begingroup$
        Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
        $endgroup$
        – BPP
        Dec 11 '18 at 20:09


















      • $begingroup$
        Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
        $endgroup$
        – BPP
        Dec 11 '18 at 20:09
















      $begingroup$
      Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
      $endgroup$
      – BPP
      Dec 11 '18 at 20:09




      $begingroup$
      Of course you can use the cosine law in triangle $ABD$ if you want to use $cos 2x$. If the position of $A$ and the value of $x$ aren't specified, you need the values of $b$ and $c$ to find them, unless $ABD$ is a right triangle.
      $endgroup$
      – BPP
      Dec 11 '18 at 20:09











      0












      $begingroup$

      To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
      $$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
        $$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
          $$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$






          share|cite|improve this answer









          $endgroup$



          To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:
          $$x = frac{-b pm sqrt{b^2 - 4ac} }{2a}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 10:57









          HawkinsCivilHawkinsCivil

          14




          14























              -1












              $begingroup$

              $cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.






              share|cite|improve this answer











              $endgroup$


















                -1












                $begingroup$

                $cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.






                share|cite|improve this answer











                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  $cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.






                  share|cite|improve this answer











                  $endgroup$



                  $cos(x)neq frac{2}{3}$ because the angle $widehat{ACD} neq frac{pi}{2}$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 11 '18 at 17:29









                  Davide Giraudo

                  127k16151264




                  127k16151264










                  answered Dec 11 '18 at 16:49









                  M.gaussM.gauss

                  91




                  91






























                      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%2f3035443%2fhow-to-use-double-angle-identities-to-find-length-d%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

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                      Read more