How do I construct this triangle












1












$begingroup$


I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.
enter image description here




  1. Is it possible to construct it without trigonometry or analytical geometry?

  2. Is it possible to show that it cannot be done without trigonometry or analytical geometry?

  3. (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?


Note 1: In the actual drawing, $c=6$, but that does not matter, obviously.
Note 2: I am not looking for LaTeX help.






geometry trigonometry triangle







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What concrete is given in the exercise?
    $endgroup$
    – Dr. Sonnhard Graubner
    2 hours ago










  • $begingroup$
    Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
    $endgroup$
    – Henry
    2 hours ago










  • $begingroup$
    And $$angle {CEB}=120^{circ}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    2 hours ago










  • $begingroup$
    @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
    $endgroup$
    – blackened
    2 hours ago










  • $begingroup$
    "figure not to scale"...
    $endgroup$
    – David Mitra
    2 hours ago
















1












$begingroup$


I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.
enter image description here




  1. Is it possible to construct it without trigonometry or analytical geometry?

  2. Is it possible to show that it cannot be done without trigonometry or analytical geometry?

  3. (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?


Note 1: In the actual drawing, $c=6$, but that does not matter, obviously.
Note 2: I am not looking for LaTeX help.






geometry trigonometry triangle







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What concrete is given in the exercise?
    $endgroup$
    – Dr. Sonnhard Graubner
    2 hours ago










  • $begingroup$
    Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
    $endgroup$
    – Henry
    2 hours ago










  • $begingroup$
    And $$angle {CEB}=120^{circ}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    2 hours ago










  • $begingroup$
    @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
    $endgroup$
    – blackened
    2 hours ago










  • $begingroup$
    "figure not to scale"...
    $endgroup$
    – David Mitra
    2 hours ago














1












1








1





$begingroup$


I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.
enter image description here




  1. Is it possible to construct it without trigonometry or analytical geometry?

  2. Is it possible to show that it cannot be done without trigonometry or analytical geometry?

  3. (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?


Note 1: In the actual drawing, $c=6$, but that does not matter, obviously.
Note 2: I am not looking for LaTeX help.






geometry trigonometry triangle







share|cite|improve this question











$endgroup$




I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.
enter image description here




  1. Is it possible to construct it without trigonometry or analytical geometry?

  2. Is it possible to show that it cannot be done without trigonometry or analytical geometry?

  3. (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?


Note 1: In the actual drawing, $c=6$, but that does not matter, obviously.
Note 2: I am not looking for LaTeX help.






geometry trigonometry triangle




geometry trigonometry triangle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 16 mins ago









Martin Sleziak

44.7k9117272




44.7k9117272










asked 2 hours ago









blackenedblackened

327211




327211








  • 1




    $begingroup$
    What concrete is given in the exercise?
    $endgroup$
    – Dr. Sonnhard Graubner
    2 hours ago










  • $begingroup$
    Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
    $endgroup$
    – Henry
    2 hours ago










  • $begingroup$
    And $$angle {CEB}=120^{circ}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    2 hours ago










  • $begingroup$
    @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
    $endgroup$
    – blackened
    2 hours ago










  • $begingroup$
    "figure not to scale"...
    $endgroup$
    – David Mitra
    2 hours ago














  • 1




    $begingroup$
    What concrete is given in the exercise?
    $endgroup$
    – Dr. Sonnhard Graubner
    2 hours ago










  • $begingroup$
    Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
    $endgroup$
    – Henry
    2 hours ago










  • $begingroup$
    And $$angle {CEB}=120^{circ}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    2 hours ago










  • $begingroup$
    @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
    $endgroup$
    – blackened
    2 hours ago










  • $begingroup$
    "figure not to scale"...
    $endgroup$
    – David Mitra
    2 hours ago








1




1




$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
2 hours ago




$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
2 hours ago












$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
2 hours ago




$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
2 hours ago












$begingroup$
And $$angle {CEB}=120^{circ}$$
$endgroup$
– Dr. Sonnhard Graubner
2 hours ago




$begingroup$
And $$angle {CEB}=120^{circ}$$
$endgroup$
– Dr. Sonnhard Graubner
2 hours ago












$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
2 hours ago




$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
2 hours ago












$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
2 hours ago




$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
2 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:



    enter image description here



    Now, recall that, in a triangle, smaller angles are opposite smaller sides.



    $$begin{align}
    triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
    triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
    end{align}$$



    Consequently, $2p < q + r$, which contradicts the desired condition. $square$






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
      For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
        For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
          For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.






          share|cite|improve this answer









          $endgroup$



          This triangle cannot be constructed for any $phi$. If we assume $phi$ can be any value and still have the triangle be constructed, we can see that length $BE = 2 cdot a$ as $phi to 60^circ$. Then applying the sine rule on lengths BE and AE gives: $$frac{2cdot a}{sin{(phi + 60^circ)}}=frac{a}{sin{(60^circ - phi)}}$$
          For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt{3} cdot a$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $AB^2+BC^2=AC^2$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Peter ForemanPeter Foreman

          3727




          3727























              2












              $begingroup$

              As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:



              enter image description here



              Now, recall that, in a triangle, smaller angles are opposite smaller sides.



              $$begin{align}
              triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
              triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
              end{align}$$



              Consequently, $2p < q + r$, which contradicts the desired condition. $square$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:



                enter image description here



                Now, recall that, in a triangle, smaller angles are opposite smaller sides.



                $$begin{align}
                triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
                triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
                end{align}$$



                Consequently, $2p < q + r$, which contradicts the desired condition. $square$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:



                  enter image description here



                  Now, recall that, in a triangle, smaller angles are opposite smaller sides.



                  $$begin{align}
                  triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
                  triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
                  end{align}$$



                  Consequently, $2p < q + r$, which contradicts the desired condition. $square$






                  share|cite|improve this answer











                  $endgroup$



                  As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overline{BC}$ such that $triangle ABEcong triangle A^prime BE$, as shown:



                  enter image description here



                  Now, recall that, in a triangle, smaller angles are opposite smaller sides.



                  $$begin{align}
                  triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
                  triangle A^prime CE: quad phantom{60^circ-;}phi < 60^circ &quadimpliesquad p < r
                  end{align}$$



                  Consequently, $2p < q + r$, which contradicts the desired condition. $square$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 1 hour ago

























                  answered 1 hour ago









                  BlueBlue

                  47.8k870152




                  47.8k870152






























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