Find the length of a triangle's side












6












$begingroup$


I have the following triangle and I'm supposed to find the value of x.



enter image description here



First thought that came to mind is to use the the following tangent equation



$$tan(y)=frac{x}{27}$$ and $$tan(19+y) = frac{2x}{27}$$ which implies that $$tan(19+y) =2tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.



I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).



Edit: Using the fact $tan(a+b) = dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ yields the following quadratic equation $$2tan(19)tan^2(y)-tan(y)+tan(19) = 0$$which gives you that $tan(y) = 0.890833$ or $tan(y)= 0.561272$ which means $x = 27times0.890833$ or $x= 27times 0.561272$.



So the question is now which $x$ should I pick and why? and is this more complicated than it should be?






geometry proof-verification proof-explanation triangle alternative-proof







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
    $endgroup$
    – poetasis
    Feb 10 at 16:39










  • $begingroup$
    Why do you assume there is only one solution?
    $endgroup$
    – fleablood
    Feb 10 at 17:46
















6












$begingroup$


I have the following triangle and I'm supposed to find the value of x.



enter image description here



First thought that came to mind is to use the the following tangent equation



$$tan(y)=frac{x}{27}$$ and $$tan(19+y) = frac{2x}{27}$$ which implies that $$tan(19+y) =2tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.



I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).



Edit: Using the fact $tan(a+b) = dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ yields the following quadratic equation $$2tan(19)tan^2(y)-tan(y)+tan(19) = 0$$which gives you that $tan(y) = 0.890833$ or $tan(y)= 0.561272$ which means $x = 27times0.890833$ or $x= 27times 0.561272$.



So the question is now which $x$ should I pick and why? and is this more complicated than it should be?






geometry proof-verification proof-explanation triangle alternative-proof







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
    $endgroup$
    – poetasis
    Feb 10 at 16:39










  • $begingroup$
    Why do you assume there is only one solution?
    $endgroup$
    – fleablood
    Feb 10 at 17:46














6












6








6





$begingroup$


I have the following triangle and I'm supposed to find the value of x.



enter image description here



First thought that came to mind is to use the the following tangent equation



$$tan(y)=frac{x}{27}$$ and $$tan(19+y) = frac{2x}{27}$$ which implies that $$tan(19+y) =2tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.



I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).



Edit: Using the fact $tan(a+b) = dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ yields the following quadratic equation $$2tan(19)tan^2(y)-tan(y)+tan(19) = 0$$which gives you that $tan(y) = 0.890833$ or $tan(y)= 0.561272$ which means $x = 27times0.890833$ or $x= 27times 0.561272$.



So the question is now which $x$ should I pick and why? and is this more complicated than it should be?






geometry proof-verification proof-explanation triangle alternative-proof







share|cite|improve this question











$endgroup$




I have the following triangle and I'm supposed to find the value of x.



enter image description here



First thought that came to mind is to use the the following tangent equation



$$tan(y)=frac{x}{27}$$ and $$tan(19+y) = frac{2x}{27}$$ which implies that $$tan(19+y) =2tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.



I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).



Edit: Using the fact $tan(a+b) = dfrac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ yields the following quadratic equation $$2tan(19)tan^2(y)-tan(y)+tan(19) = 0$$which gives you that $tan(y) = 0.890833$ or $tan(y)= 0.561272$ which means $x = 27times0.890833$ or $x= 27times 0.561272$.



So the question is now which $x$ should I pick and why? and is this more complicated than it should be?






geometry proof-verification proof-explanation triangle alternative-proof




geometry proof-verification proof-explanation triangle alternative-proof






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 10 at 16:24









Anirban Niloy

8271218




8271218










asked Feb 10 at 15:24









unknown1unknown1

747




747












  • $begingroup$
    Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
    $endgroup$
    – poetasis
    Feb 10 at 16:39










  • $begingroup$
    Why do you assume there is only one solution?
    $endgroup$
    – fleablood
    Feb 10 at 17:46


















  • $begingroup$
    Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
    $endgroup$
    – poetasis
    Feb 10 at 16:39










  • $begingroup$
    Why do you assume there is only one solution?
    $endgroup$
    – fleablood
    Feb 10 at 17:46
















$begingroup$
Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
$endgroup$
– poetasis
Feb 10 at 16:39




$begingroup$
Do you $know$ that $x$ is half the altitude of the triangle or did you simply infer it from the picture?
$endgroup$
– poetasis
Feb 10 at 16:39












$begingroup$
Why do you assume there is only one solution?
$endgroup$
– fleablood
Feb 10 at 17:46




$begingroup$
Why do you assume there is only one solution?
$endgroup$
– fleablood
Feb 10 at 17:46










2 Answers
2






active

oldest

votes


















3












$begingroup$

As you noticed $2tan(y)=tan(19+y)$. Hence
$$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
This gives $$x=15.15...lor x=24.05...$$



Now, as you can see in the following image (made with geogebra), both values are correct:



enter image description here






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I think you can just use the sinus rule and pythagorean theorem, then:
    $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
    And we see:
    $$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
    Thus, if we substitute:
    $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
    Solving for $x$:
    $$x=15.15...lor x=24.05...$$
    ... witch is in compliance with your answer.



    PS.
    Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.






    share|cite|improve this answer











    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      3












      $begingroup$

      As you noticed $2tan(y)=tan(19+y)$. Hence
      $$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
      This gives $$x=15.15...lor x=24.05...$$



      Now, as you can see in the following image (made with geogebra), both values are correct:



      enter image description here






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        As you noticed $2tan(y)=tan(19+y)$. Hence
        $$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
        This gives $$x=15.15...lor x=24.05...$$



        Now, as you can see in the following image (made with geogebra), both values are correct:



        enter image description here






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          As you noticed $2tan(y)=tan(19+y)$. Hence
          $$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
          This gives $$x=15.15...lor x=24.05...$$



          Now, as you can see in the following image (made with geogebra), both values are correct:



          enter image description here






          share|cite|improve this answer









          $endgroup$



          As you noticed $2tan(y)=tan(19+y)$. Hence
          $$2tan(y)=frac{2x}{27}=tan(19+y)=frac{tan(19)+tan(y)}{1-tan(y)tan(19)}=frac{tan(19)+frac{x}{27}}{1-tan(19)·frac{x}{27}}=frac{27tan(19)+x}{27-tan(19)x}$$ $$iff frac{2x}{27}=frac{27tan(19)+x}{27-tan(19)x}iff 54x-2tan(19)·x^2=729tan(19)+27x$$ $$iff 2tan(19)·x^2-27x+729tan(19)=0$$
          This gives $$x=15.15...lor x=24.05...$$



          Now, as you can see in the following image (made with geogebra), both values are correct:



          enter image description here







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 10 at 17:03









          Dr. MathvaDr. Mathva

          2,271526




          2,271526























              1












              $begingroup$

              I think you can just use the sinus rule and pythagorean theorem, then:
              $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
              And we see:
              $$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
              Thus, if we substitute:
              $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
              Solving for $x$:
              $$x=15.15...lor x=24.05...$$
              ... witch is in compliance with your answer.



              PS.
              Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                I think you can just use the sinus rule and pythagorean theorem, then:
                $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
                And we see:
                $$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
                Thus, if we substitute:
                $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
                Solving for $x$:
                $$x=15.15...lor x=24.05...$$
                ... witch is in compliance with your answer.



                PS.
                Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I think you can just use the sinus rule and pythagorean theorem, then:
                  $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
                  And we see:
                  $$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
                  Thus, if we substitute:
                  $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
                  Solving for $x$:
                  $$x=15.15...lor x=24.05...$$
                  ... witch is in compliance with your answer.



                  PS.
                  Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.






                  share|cite|improve this answer











                  $endgroup$



                  I think you can just use the sinus rule and pythagorean theorem, then:
                  $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}}{sin(71-y)}$$
                  And we see:
                  $$sin(71-y)=frac{27}{sqrt{4x^2+27^2}}$$
                  Thus, if we substitute:
                  $$frac{x}{sin(19)} = frac{sqrt{x^2+27^2}sqrt{4x^2+27^2}}{27}$$
                  Solving for $x$:
                  $$x=15.15...lor x=24.05...$$
                  ... witch is in compliance with your answer.



                  PS.
                  Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 10 at 17:16

























                  answered Feb 10 at 16:40









                  MaxMax

                  564317




                  564317






























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