How do I construct this triangle [closed]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












2












$begingroup$


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.










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closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn Jan 20 at 22:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    $begingroup$
    What concrete is given in the exercise?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 20 at 10:13










  • $begingroup$
    Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
    $endgroup$
    – Henry
    Jan 20 at 10:15










  • $begingroup$
    And $$angle CEB=120^circ$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 20 at 10:18










  • $begingroup$
    @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
    $endgroup$
    – blackened
    Jan 20 at 10:20






  • 1




    $begingroup$
    "figure not to scale"...
    $endgroup$
    – David Mitra
    Jan 20 at 10:21















2












$begingroup$


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.










share|cite|improve this question











$endgroup$



closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn Jan 20 at 22:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    $begingroup$
    What concrete is given in the exercise?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 20 at 10:13










  • $begingroup$
    Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
    $endgroup$
    – Henry
    Jan 20 at 10:15










  • $begingroup$
    And $$angle CEB=120^circ$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 20 at 10:18










  • $begingroup$
    @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
    $endgroup$
    – blackened
    Jan 20 at 10:20






  • 1




    $begingroup$
    "figure not to scale"...
    $endgroup$
    – David Mitra
    Jan 20 at 10:21













2












2








2





$begingroup$


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.










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






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share|cite|improve this question













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edited Jan 20 at 12:52









Martin Sleziak

44.8k9118272




44.8k9118272










asked Jan 20 at 10:10









blackenedblackened

332211




332211




closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn Jan 20 at 22:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn Jan 20 at 22:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    $begingroup$
    What concrete is given in the exercise?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 20 at 10:13










  • $begingroup$
    Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
    $endgroup$
    – Henry
    Jan 20 at 10:15










  • $begingroup$
    And $$angle CEB=120^circ$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 20 at 10:18










  • $begingroup$
    @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
    $endgroup$
    – blackened
    Jan 20 at 10:20






  • 1




    $begingroup$
    "figure not to scale"...
    $endgroup$
    – David Mitra
    Jan 20 at 10:21












  • 1




    $begingroup$
    What concrete is given in the exercise?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 20 at 10:13










  • $begingroup$
    Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
    $endgroup$
    – Henry
    Jan 20 at 10:15










  • $begingroup$
    And $$angle CEB=120^circ$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 20 at 10:18










  • $begingroup$
    @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
    $endgroup$
    – blackened
    Jan 20 at 10:20






  • 1




    $begingroup$
    "figure not to scale"...
    $endgroup$
    – David Mitra
    Jan 20 at 10:21







1




1




$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:13




$begingroup$
What concrete is given in the exercise?
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:13












$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
Jan 20 at 10:15




$begingroup$
Are you looking for $LaTeX$ help or geometrical help such as $psi = 60^circ-phi$
$endgroup$
– Henry
Jan 20 at 10:15












$begingroup$
And $$angle CEB=120^circ$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:18




$begingroup$
And $$angle CEB=120^circ$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 20 at 10:18












$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
Jan 20 at 10:20




$begingroup$
@Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met.
$endgroup$
– blackened
Jan 20 at 10:20




1




1




$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
Jan 20 at 10:21




$begingroup$
"figure not to scale"...
$endgroup$
– David Mitra
Jan 20 at 10:21










2 Answers
2






active

oldest

votes


















5












$begingroup$

As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overlineBC$ 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.



$$beginalign
triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
triangle A^prime CE: quad phantom60^circ-;phi < 60^circ &quadimpliesquad p < r
endalign$$



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






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: $$frac2cdot asin(phi + 60^circ)=fracasin(60^circ - phi)$$
    For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt3 cdot a$ 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 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overlineBC$ 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.



      $$beginalign
      triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
      triangle A^prime CE: quad phantom60^circ-;phi < 60^circ &quadimpliesquad p < r
      endalign$$



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






      share|cite|improve this answer











      $endgroup$

















        5












        $begingroup$

        As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overlineBC$ 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.



        $$beginalign
        triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
        triangle A^prime CE: quad phantom60^circ-;phi < 60^circ &quadimpliesquad p < r
        endalign$$



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






        share|cite|improve this answer











        $endgroup$















          5












          5








          5





          $begingroup$

          As noted in the comments, a little angle-chasing shows that $phi + psi = 60^circ$. As a result, we can find $A^prime$ on $overlineBC$ 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.



          $$beginalign
          triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
          triangle A^prime CE: quad phantom60^circ-;phi < 60^circ &quadimpliesquad p < r
          endalign$$



          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 $overlineBC$ 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.



          $$beginalign
          triangle ABE: quad 60^circ - phi < 60^circ &quadimpliesquad p < q \
          triangle A^prime CE: quad phantom60^circ-;phi < 60^circ &quadimpliesquad p < r
          endalign$$



          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 Jan 20 at 11:23

























          answered Jan 20 at 11:17









          BlueBlue

          48.3k870153




          48.3k870153





















              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: $$frac2cdot asin(phi + 60^circ)=fracasin(60^circ - phi)$$
              For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt3 cdot a$ 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: $$frac2cdot asin(phi + 60^circ)=fracasin(60^circ - phi)$$
                For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt3 cdot a$ 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: $$frac2cdot asin(phi + 60^circ)=fracasin(60^circ - phi)$$
                  For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt3 cdot a$ 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: $$frac2cdot asin(phi + 60^circ)=fracasin(60^circ - phi)$$
                  For which the only solutions are $a=0$ (invalid) or $phi = 30^circ$. If $phi = 30^circ$, then length $AB =sqrt3 cdot a$ 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 Jan 20 at 10:42









                  Peter ForemanPeter Foreman

                  1,13712




                  1,13712












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