Solve similar right triangles, given one's hypotenuse, the other's base, and the sum of their heights.












2












$begingroup$


I encountered this problem while trying to determine a generic equation for entasis, but this question is not about entasis.



$theta$ is wanted—given this lovely figure



enter image description here



given that the two triangles are similar, and given $a$, $b$, and $h$.



I recognize that the sum of the heights of the triangles equals $h$, and that their ratio equals the scale factor, which seems like a likely avenue, but my trigonometry and geometry are weak and I can’t figure this one out.






geometry trigonometry euclidean-geometry triangle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Seems doable in principle. If you let $mu$ be the ratio of the lower altitude to $h$, you can set up a quartic equation in $mu$.
    $endgroup$
    – amd
    Dec 21 '18 at 23:50










  • $begingroup$
    @amd So $mu = frac{text{lower altitude}}{h}$? How from there to a quartic? or, what gets raised to the fourth?
    $endgroup$
    – holomenicus
    Dec 21 '18 at 23:55










  • $begingroup$
    It’s a quartic in $mu$. Alternatively, you can set up a pair of equations involving trig function of $theta$ and then eliminate $mu$ from them.
    $endgroup$
    – amd
    Dec 22 '18 at 0:08
















2












$begingroup$


I encountered this problem while trying to determine a generic equation for entasis, but this question is not about entasis.



$theta$ is wanted—given this lovely figure



enter image description here



given that the two triangles are similar, and given $a$, $b$, and $h$.



I recognize that the sum of the heights of the triangles equals $h$, and that their ratio equals the scale factor, which seems like a likely avenue, but my trigonometry and geometry are weak and I can’t figure this one out.






geometry trigonometry euclidean-geometry triangle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Seems doable in principle. If you let $mu$ be the ratio of the lower altitude to $h$, you can set up a quartic equation in $mu$.
    $endgroup$
    – amd
    Dec 21 '18 at 23:50










  • $begingroup$
    @amd So $mu = frac{text{lower altitude}}{h}$? How from there to a quartic? or, what gets raised to the fourth?
    $endgroup$
    – holomenicus
    Dec 21 '18 at 23:55










  • $begingroup$
    It’s a quartic in $mu$. Alternatively, you can set up a pair of equations involving trig function of $theta$ and then eliminate $mu$ from them.
    $endgroup$
    – amd
    Dec 22 '18 at 0:08














2












2








2


1



$begingroup$


I encountered this problem while trying to determine a generic equation for entasis, but this question is not about entasis.



$theta$ is wanted—given this lovely figure



enter image description here



given that the two triangles are similar, and given $a$, $b$, and $h$.



I recognize that the sum of the heights of the triangles equals $h$, and that their ratio equals the scale factor, which seems like a likely avenue, but my trigonometry and geometry are weak and I can’t figure this one out.






geometry trigonometry euclidean-geometry triangle







share|cite|improve this question











$endgroup$




I encountered this problem while trying to determine a generic equation for entasis, but this question is not about entasis.



$theta$ is wanted—given this lovely figure



enter image description here



given that the two triangles are similar, and given $a$, $b$, and $h$.



I recognize that the sum of the heights of the triangles equals $h$, and that their ratio equals the scale factor, which seems like a likely avenue, but my trigonometry and geometry are weak and I can’t figure this one out.






geometry trigonometry euclidean-geometry triangle




geometry trigonometry euclidean-geometry triangle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 1:03









Blue

48.7k870156




48.7k870156










asked Dec 21 '18 at 23:37









holomenicusholomenicus

174




174












  • $begingroup$
    Seems doable in principle. If you let $mu$ be the ratio of the lower altitude to $h$, you can set up a quartic equation in $mu$.
    $endgroup$
    – amd
    Dec 21 '18 at 23:50










  • $begingroup$
    @amd So $mu = frac{text{lower altitude}}{h}$? How from there to a quartic? or, what gets raised to the fourth?
    $endgroup$
    – holomenicus
    Dec 21 '18 at 23:55










  • $begingroup$
    It’s a quartic in $mu$. Alternatively, you can set up a pair of equations involving trig function of $theta$ and then eliminate $mu$ from them.
    $endgroup$
    – amd
    Dec 22 '18 at 0:08


















  • $begingroup$
    Seems doable in principle. If you let $mu$ be the ratio of the lower altitude to $h$, you can set up a quartic equation in $mu$.
    $endgroup$
    – amd
    Dec 21 '18 at 23:50










  • $begingroup$
    @amd So $mu = frac{text{lower altitude}}{h}$? How from there to a quartic? or, what gets raised to the fourth?
    $endgroup$
    – holomenicus
    Dec 21 '18 at 23:55










  • $begingroup$
    It’s a quartic in $mu$. Alternatively, you can set up a pair of equations involving trig function of $theta$ and then eliminate $mu$ from them.
    $endgroup$
    – amd
    Dec 22 '18 at 0:08
















$begingroup$
Seems doable in principle. If you let $mu$ be the ratio of the lower altitude to $h$, you can set up a quartic equation in $mu$.
$endgroup$
– amd
Dec 21 '18 at 23:50




$begingroup$
Seems doable in principle. If you let $mu$ be the ratio of the lower altitude to $h$, you can set up a quartic equation in $mu$.
$endgroup$
– amd
Dec 21 '18 at 23:50












$begingroup$
@amd So $mu = frac{text{lower altitude}}{h}$? How from there to a quartic? or, what gets raised to the fourth?
$endgroup$
– holomenicus
Dec 21 '18 at 23:55




$begingroup$
@amd So $mu = frac{text{lower altitude}}{h}$? How from there to a quartic? or, what gets raised to the fourth?
$endgroup$
– holomenicus
Dec 21 '18 at 23:55












$begingroup$
It’s a quartic in $mu$. Alternatively, you can set up a pair of equations involving trig function of $theta$ and then eliminate $mu$ from them.
$endgroup$
– amd
Dec 22 '18 at 0:08




$begingroup$
It’s a quartic in $mu$. Alternatively, you can set up a pair of equations involving trig function of $theta$ and then eliminate $mu$ from them.
$endgroup$
– amd
Dec 22 '18 at 0:08










2 Answers
2






active

oldest

votes


















1












$begingroup$

It is pretty easy to see from the geometry of the figure that



$b cot theta + a cos theta = h, tag 1$



whence,



$b dfrac{cos theta}{sin theta} + acos theta = h; tag 2$



now using $cos^2 theta + sin^2 theta = 1$, i.e. $cos theta = pm sqrt{1 - sin^2 theta}$,



$b dfrac{sqrt{1 - sin^2 theta}}{sin theta} + a sqrt{1 - sin^2 theta} = h; tag 3$



we choose the positive sign on $pm sqrt{1 - sin^2 theta}$ since the angle $theta$ appears to be acute; next, we square:



$b^2 dfrac{1 - sin^2 theta}{sin^2 theta} + 2ab dfrac{1 - sin^2 theta}{sin theta} + a^2 (1 - sin^2 theta) = h^2; tag 4$



we multiply by $sin^2 theta$:



$b^2 (1 - sin^2 theta)+ 2ab sin theta (1 - sin^2 theta) + a^2 sin^2 theta (1 - sin^2 theta) = h^2 sin^2 theta, tag 5$



which may be written as a quartic equation in $sin theta$:



$-a^2 sin^4 theta -2ab sin^3 theta + (a^2 - b^2 - h^2)sin^2 theta + 2ab sin theta + b^2 = 0, tag 6$



or



$a^2 sin^4 theta + 2ab sin^3 theta + (h^2 - a^2 - b^2)sin^2 theta - 2ab sin theta - b^2 = 0. tag 7$



This is about as far as we can push things using elementary algebra and trigonometry. To find $sin theta$, we must solve this quartic, which may be done according to this wikipedia page.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Let $mu$ be the ratio of the lower triangle’s altitude to the total height $h$. We then have $$tantheta = {b over mu h} \ costheta = {(1-mu)h over a}.$$ Eliminate $mu$ to get the equation $$h = acostheta + bcottheta.$$ Alternatively, you can apply the Pythagorean theorem to get the equation $$(mu h)^2+b^2 = left({mu aover 1-mu}right)^2,$$ a quartic in $mu$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      For a $mu$-free solution, isolate $bcottheta$ in your $h$ equation, and multiply-through by $sintheta$ ... $$begin{align} sintheta (h-acostheta)= b costheta &quadtoquad sin^2theta (h-acostheta)^2 = b^2 cos^2theta \ &quadtoquad (1-cos^2theta) (h-acostheta)^2 = b^2 cos^2theta end{align}$$ Abbreviating $costheta$ as $k$, this gives $$a^2 k^4- 2 a h k^3+ ( h^2 - a^2 + b^2 )k^2+ 2 a h k - h^2 = 0$$ This quartic, in general, has ugly roots.
      $endgroup$
      – Blue
      Dec 22 '18 at 1:29








    • 1




      $begingroup$
      @Blue You've got that right. $ echo "solve(a^2 * k^4 - 2*a*h*k^3 + (h^2 - a^2 + b^2)*k^2 + 2*a*h*k - h^2 = 0, k);" | maxima | wc -l prints 295 lines of equation to express the roots.
      $endgroup$
      – holomenicus
      Dec 22 '18 at 1:39










    • $begingroup$
      @Blue I suspect that a quartic is unavoidable.
      $endgroup$
      – amd
      Dec 22 '18 at 2:32










    • $begingroup$
      @amd: Agreed. By the way, under the substitution $sintheta = 2t/(1+t^2)$, $costheta = (1-t^2)/(1+t^2)$ in the $h$ equation, we get this quartic $$b t^4 + 2 (a+h) t^3 - 2 (a-h) t - b = 0$$ The roots are still ugly, but at least the coefficients are nicer. :)
      $endgroup$
      – Blue
      Dec 22 '18 at 2:38











    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%2f3049005%2fsolve-similar-right-triangles-given-ones-hypotenuse-the-others-base-and-the%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    It is pretty easy to see from the geometry of the figure that



    $b cot theta + a cos theta = h, tag 1$



    whence,



    $b dfrac{cos theta}{sin theta} + acos theta = h; tag 2$



    now using $cos^2 theta + sin^2 theta = 1$, i.e. $cos theta = pm sqrt{1 - sin^2 theta}$,



    $b dfrac{sqrt{1 - sin^2 theta}}{sin theta} + a sqrt{1 - sin^2 theta} = h; tag 3$



    we choose the positive sign on $pm sqrt{1 - sin^2 theta}$ since the angle $theta$ appears to be acute; next, we square:



    $b^2 dfrac{1 - sin^2 theta}{sin^2 theta} + 2ab dfrac{1 - sin^2 theta}{sin theta} + a^2 (1 - sin^2 theta) = h^2; tag 4$



    we multiply by $sin^2 theta$:



    $b^2 (1 - sin^2 theta)+ 2ab sin theta (1 - sin^2 theta) + a^2 sin^2 theta (1 - sin^2 theta) = h^2 sin^2 theta, tag 5$



    which may be written as a quartic equation in $sin theta$:



    $-a^2 sin^4 theta -2ab sin^3 theta + (a^2 - b^2 - h^2)sin^2 theta + 2ab sin theta + b^2 = 0, tag 6$



    or



    $a^2 sin^4 theta + 2ab sin^3 theta + (h^2 - a^2 - b^2)sin^2 theta - 2ab sin theta - b^2 = 0. tag 7$



    This is about as far as we can push things using elementary algebra and trigonometry. To find $sin theta$, we must solve this quartic, which may be done according to this wikipedia page.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      It is pretty easy to see from the geometry of the figure that



      $b cot theta + a cos theta = h, tag 1$



      whence,



      $b dfrac{cos theta}{sin theta} + acos theta = h; tag 2$



      now using $cos^2 theta + sin^2 theta = 1$, i.e. $cos theta = pm sqrt{1 - sin^2 theta}$,



      $b dfrac{sqrt{1 - sin^2 theta}}{sin theta} + a sqrt{1 - sin^2 theta} = h; tag 3$



      we choose the positive sign on $pm sqrt{1 - sin^2 theta}$ since the angle $theta$ appears to be acute; next, we square:



      $b^2 dfrac{1 - sin^2 theta}{sin^2 theta} + 2ab dfrac{1 - sin^2 theta}{sin theta} + a^2 (1 - sin^2 theta) = h^2; tag 4$



      we multiply by $sin^2 theta$:



      $b^2 (1 - sin^2 theta)+ 2ab sin theta (1 - sin^2 theta) + a^2 sin^2 theta (1 - sin^2 theta) = h^2 sin^2 theta, tag 5$



      which may be written as a quartic equation in $sin theta$:



      $-a^2 sin^4 theta -2ab sin^3 theta + (a^2 - b^2 - h^2)sin^2 theta + 2ab sin theta + b^2 = 0, tag 6$



      or



      $a^2 sin^4 theta + 2ab sin^3 theta + (h^2 - a^2 - b^2)sin^2 theta - 2ab sin theta - b^2 = 0. tag 7$



      This is about as far as we can push things using elementary algebra and trigonometry. To find $sin theta$, we must solve this quartic, which may be done according to this wikipedia page.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        It is pretty easy to see from the geometry of the figure that



        $b cot theta + a cos theta = h, tag 1$



        whence,



        $b dfrac{cos theta}{sin theta} + acos theta = h; tag 2$



        now using $cos^2 theta + sin^2 theta = 1$, i.e. $cos theta = pm sqrt{1 - sin^2 theta}$,



        $b dfrac{sqrt{1 - sin^2 theta}}{sin theta} + a sqrt{1 - sin^2 theta} = h; tag 3$



        we choose the positive sign on $pm sqrt{1 - sin^2 theta}$ since the angle $theta$ appears to be acute; next, we square:



        $b^2 dfrac{1 - sin^2 theta}{sin^2 theta} + 2ab dfrac{1 - sin^2 theta}{sin theta} + a^2 (1 - sin^2 theta) = h^2; tag 4$



        we multiply by $sin^2 theta$:



        $b^2 (1 - sin^2 theta)+ 2ab sin theta (1 - sin^2 theta) + a^2 sin^2 theta (1 - sin^2 theta) = h^2 sin^2 theta, tag 5$



        which may be written as a quartic equation in $sin theta$:



        $-a^2 sin^4 theta -2ab sin^3 theta + (a^2 - b^2 - h^2)sin^2 theta + 2ab sin theta + b^2 = 0, tag 6$



        or



        $a^2 sin^4 theta + 2ab sin^3 theta + (h^2 - a^2 - b^2)sin^2 theta - 2ab sin theta - b^2 = 0. tag 7$



        This is about as far as we can push things using elementary algebra and trigonometry. To find $sin theta$, we must solve this quartic, which may be done according to this wikipedia page.






        share|cite|improve this answer











        $endgroup$



        It is pretty easy to see from the geometry of the figure that



        $b cot theta + a cos theta = h, tag 1$



        whence,



        $b dfrac{cos theta}{sin theta} + acos theta = h; tag 2$



        now using $cos^2 theta + sin^2 theta = 1$, i.e. $cos theta = pm sqrt{1 - sin^2 theta}$,



        $b dfrac{sqrt{1 - sin^2 theta}}{sin theta} + a sqrt{1 - sin^2 theta} = h; tag 3$



        we choose the positive sign on $pm sqrt{1 - sin^2 theta}$ since the angle $theta$ appears to be acute; next, we square:



        $b^2 dfrac{1 - sin^2 theta}{sin^2 theta} + 2ab dfrac{1 - sin^2 theta}{sin theta} + a^2 (1 - sin^2 theta) = h^2; tag 4$



        we multiply by $sin^2 theta$:



        $b^2 (1 - sin^2 theta)+ 2ab sin theta (1 - sin^2 theta) + a^2 sin^2 theta (1 - sin^2 theta) = h^2 sin^2 theta, tag 5$



        which may be written as a quartic equation in $sin theta$:



        $-a^2 sin^4 theta -2ab sin^3 theta + (a^2 - b^2 - h^2)sin^2 theta + 2ab sin theta + b^2 = 0, tag 6$



        or



        $a^2 sin^4 theta + 2ab sin^3 theta + (h^2 - a^2 - b^2)sin^2 theta - 2ab sin theta - b^2 = 0. tag 7$



        This is about as far as we can push things using elementary algebra and trigonometry. To find $sin theta$, we must solve this quartic, which may be done according to this wikipedia page.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 22 '18 at 3:20

























        answered Dec 22 '18 at 3:15









        Robert LewisRobert Lewis

        47.5k23067




        47.5k23067























            2












            $begingroup$

            Let $mu$ be the ratio of the lower triangle’s altitude to the total height $h$. We then have $$tantheta = {b over mu h} \ costheta = {(1-mu)h over a}.$$ Eliminate $mu$ to get the equation $$h = acostheta + bcottheta.$$ Alternatively, you can apply the Pythagorean theorem to get the equation $$(mu h)^2+b^2 = left({mu aover 1-mu}right)^2,$$ a quartic in $mu$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              For a $mu$-free solution, isolate $bcottheta$ in your $h$ equation, and multiply-through by $sintheta$ ... $$begin{align} sintheta (h-acostheta)= b costheta &quadtoquad sin^2theta (h-acostheta)^2 = b^2 cos^2theta \ &quadtoquad (1-cos^2theta) (h-acostheta)^2 = b^2 cos^2theta end{align}$$ Abbreviating $costheta$ as $k$, this gives $$a^2 k^4- 2 a h k^3+ ( h^2 - a^2 + b^2 )k^2+ 2 a h k - h^2 = 0$$ This quartic, in general, has ugly roots.
              $endgroup$
              – Blue
              Dec 22 '18 at 1:29








            • 1




              $begingroup$
              @Blue You've got that right. $ echo "solve(a^2 * k^4 - 2*a*h*k^3 + (h^2 - a^2 + b^2)*k^2 + 2*a*h*k - h^2 = 0, k);" | maxima | wc -l prints 295 lines of equation to express the roots.
              $endgroup$
              – holomenicus
              Dec 22 '18 at 1:39










            • $begingroup$
              @Blue I suspect that a quartic is unavoidable.
              $endgroup$
              – amd
              Dec 22 '18 at 2:32










            • $begingroup$
              @amd: Agreed. By the way, under the substitution $sintheta = 2t/(1+t^2)$, $costheta = (1-t^2)/(1+t^2)$ in the $h$ equation, we get this quartic $$b t^4 + 2 (a+h) t^3 - 2 (a-h) t - b = 0$$ The roots are still ugly, but at least the coefficients are nicer. :)
              $endgroup$
              – Blue
              Dec 22 '18 at 2:38
















            2












            $begingroup$

            Let $mu$ be the ratio of the lower triangle’s altitude to the total height $h$. We then have $$tantheta = {b over mu h} \ costheta = {(1-mu)h over a}.$$ Eliminate $mu$ to get the equation $$h = acostheta + bcottheta.$$ Alternatively, you can apply the Pythagorean theorem to get the equation $$(mu h)^2+b^2 = left({mu aover 1-mu}right)^2,$$ a quartic in $mu$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              For a $mu$-free solution, isolate $bcottheta$ in your $h$ equation, and multiply-through by $sintheta$ ... $$begin{align} sintheta (h-acostheta)= b costheta &quadtoquad sin^2theta (h-acostheta)^2 = b^2 cos^2theta \ &quadtoquad (1-cos^2theta) (h-acostheta)^2 = b^2 cos^2theta end{align}$$ Abbreviating $costheta$ as $k$, this gives $$a^2 k^4- 2 a h k^3+ ( h^2 - a^2 + b^2 )k^2+ 2 a h k - h^2 = 0$$ This quartic, in general, has ugly roots.
              $endgroup$
              – Blue
              Dec 22 '18 at 1:29








            • 1




              $begingroup$
              @Blue You've got that right. $ echo "solve(a^2 * k^4 - 2*a*h*k^3 + (h^2 - a^2 + b^2)*k^2 + 2*a*h*k - h^2 = 0, k);" | maxima | wc -l prints 295 lines of equation to express the roots.
              $endgroup$
              – holomenicus
              Dec 22 '18 at 1:39










            • $begingroup$
              @Blue I suspect that a quartic is unavoidable.
              $endgroup$
              – amd
              Dec 22 '18 at 2:32










            • $begingroup$
              @amd: Agreed. By the way, under the substitution $sintheta = 2t/(1+t^2)$, $costheta = (1-t^2)/(1+t^2)$ in the $h$ equation, we get this quartic $$b t^4 + 2 (a+h) t^3 - 2 (a-h) t - b = 0$$ The roots are still ugly, but at least the coefficients are nicer. :)
              $endgroup$
              – Blue
              Dec 22 '18 at 2:38














            2












            2








            2





            $begingroup$

            Let $mu$ be the ratio of the lower triangle’s altitude to the total height $h$. We then have $$tantheta = {b over mu h} \ costheta = {(1-mu)h over a}.$$ Eliminate $mu$ to get the equation $$h = acostheta + bcottheta.$$ Alternatively, you can apply the Pythagorean theorem to get the equation $$(mu h)^2+b^2 = left({mu aover 1-mu}right)^2,$$ a quartic in $mu$.






            share|cite|improve this answer









            $endgroup$



            Let $mu$ be the ratio of the lower triangle’s altitude to the total height $h$. We then have $$tantheta = {b over mu h} \ costheta = {(1-mu)h over a}.$$ Eliminate $mu$ to get the equation $$h = acostheta + bcottheta.$$ Alternatively, you can apply the Pythagorean theorem to get the equation $$(mu h)^2+b^2 = left({mu aover 1-mu}right)^2,$$ a quartic in $mu$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 22 '18 at 0:15









            amdamd

            30.7k21050




            30.7k21050












            • $begingroup$
              For a $mu$-free solution, isolate $bcottheta$ in your $h$ equation, and multiply-through by $sintheta$ ... $$begin{align} sintheta (h-acostheta)= b costheta &quadtoquad sin^2theta (h-acostheta)^2 = b^2 cos^2theta \ &quadtoquad (1-cos^2theta) (h-acostheta)^2 = b^2 cos^2theta end{align}$$ Abbreviating $costheta$ as $k$, this gives $$a^2 k^4- 2 a h k^3+ ( h^2 - a^2 + b^2 )k^2+ 2 a h k - h^2 = 0$$ This quartic, in general, has ugly roots.
              $endgroup$
              – Blue
              Dec 22 '18 at 1:29








            • 1




              $begingroup$
              @Blue You've got that right. $ echo "solve(a^2 * k^4 - 2*a*h*k^3 + (h^2 - a^2 + b^2)*k^2 + 2*a*h*k - h^2 = 0, k);" | maxima | wc -l prints 295 lines of equation to express the roots.
              $endgroup$
              – holomenicus
              Dec 22 '18 at 1:39










            • $begingroup$
              @Blue I suspect that a quartic is unavoidable.
              $endgroup$
              – amd
              Dec 22 '18 at 2:32










            • $begingroup$
              @amd: Agreed. By the way, under the substitution $sintheta = 2t/(1+t^2)$, $costheta = (1-t^2)/(1+t^2)$ in the $h$ equation, we get this quartic $$b t^4 + 2 (a+h) t^3 - 2 (a-h) t - b = 0$$ The roots are still ugly, but at least the coefficients are nicer. :)
              $endgroup$
              – Blue
              Dec 22 '18 at 2:38


















            • $begingroup$
              For a $mu$-free solution, isolate $bcottheta$ in your $h$ equation, and multiply-through by $sintheta$ ... $$begin{align} sintheta (h-acostheta)= b costheta &quadtoquad sin^2theta (h-acostheta)^2 = b^2 cos^2theta \ &quadtoquad (1-cos^2theta) (h-acostheta)^2 = b^2 cos^2theta end{align}$$ Abbreviating $costheta$ as $k$, this gives $$a^2 k^4- 2 a h k^3+ ( h^2 - a^2 + b^2 )k^2+ 2 a h k - h^2 = 0$$ This quartic, in general, has ugly roots.
              $endgroup$
              – Blue
              Dec 22 '18 at 1:29








            • 1




              $begingroup$
              @Blue You've got that right. $ echo "solve(a^2 * k^4 - 2*a*h*k^3 + (h^2 - a^2 + b^2)*k^2 + 2*a*h*k - h^2 = 0, k);" | maxima | wc -l prints 295 lines of equation to express the roots.
              $endgroup$
              – holomenicus
              Dec 22 '18 at 1:39










            • $begingroup$
              @Blue I suspect that a quartic is unavoidable.
              $endgroup$
              – amd
              Dec 22 '18 at 2:32










            • $begingroup$
              @amd: Agreed. By the way, under the substitution $sintheta = 2t/(1+t^2)$, $costheta = (1-t^2)/(1+t^2)$ in the $h$ equation, we get this quartic $$b t^4 + 2 (a+h) t^3 - 2 (a-h) t - b = 0$$ The roots are still ugly, but at least the coefficients are nicer. :)
              $endgroup$
              – Blue
              Dec 22 '18 at 2:38
















            $begingroup$
            For a $mu$-free solution, isolate $bcottheta$ in your $h$ equation, and multiply-through by $sintheta$ ... $$begin{align} sintheta (h-acostheta)= b costheta &quadtoquad sin^2theta (h-acostheta)^2 = b^2 cos^2theta \ &quadtoquad (1-cos^2theta) (h-acostheta)^2 = b^2 cos^2theta end{align}$$ Abbreviating $costheta$ as $k$, this gives $$a^2 k^4- 2 a h k^3+ ( h^2 - a^2 + b^2 )k^2+ 2 a h k - h^2 = 0$$ This quartic, in general, has ugly roots.
            $endgroup$
            – Blue
            Dec 22 '18 at 1:29






            $begingroup$
            For a $mu$-free solution, isolate $bcottheta$ in your $h$ equation, and multiply-through by $sintheta$ ... $$begin{align} sintheta (h-acostheta)= b costheta &quadtoquad sin^2theta (h-acostheta)^2 = b^2 cos^2theta \ &quadtoquad (1-cos^2theta) (h-acostheta)^2 = b^2 cos^2theta end{align}$$ Abbreviating $costheta$ as $k$, this gives $$a^2 k^4- 2 a h k^3+ ( h^2 - a^2 + b^2 )k^2+ 2 a h k - h^2 = 0$$ This quartic, in general, has ugly roots.
            $endgroup$
            – Blue
            Dec 22 '18 at 1:29






            1




            1




            $begingroup$
            @Blue You've got that right. $ echo "solve(a^2 * k^4 - 2*a*h*k^3 + (h^2 - a^2 + b^2)*k^2 + 2*a*h*k - h^2 = 0, k);" | maxima | wc -l prints 295 lines of equation to express the roots.
            $endgroup$
            – holomenicus
            Dec 22 '18 at 1:39




            $begingroup$
            @Blue You've got that right. $ echo "solve(a^2 * k^4 - 2*a*h*k^3 + (h^2 - a^2 + b^2)*k^2 + 2*a*h*k - h^2 = 0, k);" | maxima | wc -l prints 295 lines of equation to express the roots.
            $endgroup$
            – holomenicus
            Dec 22 '18 at 1:39












            $begingroup$
            @Blue I suspect that a quartic is unavoidable.
            $endgroup$
            – amd
            Dec 22 '18 at 2:32




            $begingroup$
            @Blue I suspect that a quartic is unavoidable.
            $endgroup$
            – amd
            Dec 22 '18 at 2:32












            $begingroup$
            @amd: Agreed. By the way, under the substitution $sintheta = 2t/(1+t^2)$, $costheta = (1-t^2)/(1+t^2)$ in the $h$ equation, we get this quartic $$b t^4 + 2 (a+h) t^3 - 2 (a-h) t - b = 0$$ The roots are still ugly, but at least the coefficients are nicer. :)
            $endgroup$
            – Blue
            Dec 22 '18 at 2:38




            $begingroup$
            @amd: Agreed. By the way, under the substitution $sintheta = 2t/(1+t^2)$, $costheta = (1-t^2)/(1+t^2)$ in the $h$ equation, we get this quartic $$b t^4 + 2 (a+h) t^3 - 2 (a-h) t - b = 0$$ The roots are still ugly, but at least the coefficients are nicer. :)
            $endgroup$
            – Blue
            Dec 22 '18 at 2:38


















            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%2f3049005%2fsolve-similar-right-triangles-given-ones-hypotenuse-the-others-base-and-the%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

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more