Find $x$ angle in triangle












4












$begingroup$


I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.



enter image description here






geometry euclidean-geometry triangle







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  • 1




    $begingroup$
    thats trigonometry solution. not what im looking for
    $endgroup$
    – Andriy Khrystyanovich
    9 hours ago












  • $begingroup$
    @Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
    $endgroup$
    – Michael Rozenberg
    9 hours ago










  • $begingroup$
    Note: A trigonometric solution is offered in this question.
    $endgroup$
    – Blue
    9 hours ago










  • $begingroup$
    math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
    $endgroup$
    – Matteo
    9 hours ago












  • $begingroup$
    Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
    $endgroup$
    – Aretino
    8 hours ago
















4












$begingroup$


I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.



enter image description here






geometry euclidean-geometry triangle







share|cite|improve this question









New contributor




Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    thats trigonometry solution. not what im looking for
    $endgroup$
    – Andriy Khrystyanovich
    9 hours ago












  • $begingroup$
    @Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
    $endgroup$
    – Michael Rozenberg
    9 hours ago










  • $begingroup$
    Note: A trigonometric solution is offered in this question.
    $endgroup$
    – Blue
    9 hours ago










  • $begingroup$
    math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
    $endgroup$
    – Matteo
    9 hours ago












  • $begingroup$
    Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
    $endgroup$
    – Aretino
    8 hours ago














4












4








4





$begingroup$


I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.



enter image description here






geometry euclidean-geometry triangle







share|cite|improve this question









New contributor




Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.



enter image description here






geometry euclidean-geometry triangle




geometry euclidean-geometry triangle






share|cite|improve this question









New contributor




Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









TheSimpliFire

12.6k62461




12.6k62461






New contributor




Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 9 hours ago









Andriy KhrystyanovichAndriy Khrystyanovich

1234




1234




New contributor




Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Andriy Khrystyanovich is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    thats trigonometry solution. not what im looking for
    $endgroup$
    – Andriy Khrystyanovich
    9 hours ago












  • $begingroup$
    @Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
    $endgroup$
    – Michael Rozenberg
    9 hours ago










  • $begingroup$
    Note: A trigonometric solution is offered in this question.
    $endgroup$
    – Blue
    9 hours ago










  • $begingroup$
    math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
    $endgroup$
    – Matteo
    9 hours ago












  • $begingroup$
    Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
    $endgroup$
    – Aretino
    8 hours ago














  • 1




    $begingroup$
    thats trigonometry solution. not what im looking for
    $endgroup$
    – Andriy Khrystyanovich
    9 hours ago












  • $begingroup$
    @Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
    $endgroup$
    – Michael Rozenberg
    9 hours ago










  • $begingroup$
    Note: A trigonometric solution is offered in this question.
    $endgroup$
    – Blue
    9 hours ago










  • $begingroup$
    math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
    $endgroup$
    – Matteo
    9 hours ago












  • $begingroup$
    Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
    $endgroup$
    – Aretino
    8 hours ago








1




1




$begingroup$
thats trigonometry solution. not what im looking for
$endgroup$
– Andriy Khrystyanovich
9 hours ago






$begingroup$
thats trigonometry solution. not what im looking for
$endgroup$
– Andriy Khrystyanovich
9 hours ago














$begingroup$
@Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
$endgroup$
– Michael Rozenberg
9 hours ago




$begingroup$
@Blue The solution in the linked topic by trigonometry only, because the topic starter looked for trigonometric solution only. I think we need to open this topic.
$endgroup$
– Michael Rozenberg
9 hours ago












$begingroup$
Note: A trigonometric solution is offered in this question.
$endgroup$
– Blue
9 hours ago




$begingroup$
Note: A trigonometric solution is offered in this question.
$endgroup$
– Blue
9 hours ago












$begingroup$
math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
$endgroup$
– Matteo
9 hours ago






$begingroup$
math.stackexchange.com/a/3126628/480425. Here I propose a couple of nice solutions using simple Euclidean geometry.
$endgroup$
– Matteo
9 hours ago














$begingroup$
Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
$endgroup$
– Aretino
8 hours ago




$begingroup$
Possible duplicate of In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$. Where's my error?
$endgroup$
– Aretino
8 hours ago










4 Answers
4






active

oldest

votes


















9












$begingroup$

Construct an equilateral triangle such that its sides are equal to the base of the main triangle.
enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
    $endgroup$
    – greedoid
    8 hours ago












  • $begingroup$
    @greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
    $endgroup$
    – Seyed
    8 hours ago












  • $begingroup$
    Yes, that is correct if you want another upvote.
    $endgroup$
    – greedoid
    8 hours ago






  • 2




    $begingroup$
    @greedoid, Thanks for your advise.
    $endgroup$
    – Seyed
    8 hours ago



















6












$begingroup$

And my second solution is as follow:
enter image description here






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.



    Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.



    Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$



    Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
    $$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
    But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$



    Thus, $$B'C'=B'K=AD=BC,$$ which says that
    $$Bequiv B'$$ and $$Cequiv C'.$$
    Id est,
    $$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice as always +1
      $endgroup$
      – greedoid
      8 hours ago





















    4












    $begingroup$

    enter image description here



    construct triangle $Delta BCE$ congruent to $Delta ADB$.



    so $AB = BE$, $angle ABE = 80° - 20° = 60°$



    Thus triangle $Delta ABE$ is equilateral.



    $AB = AE = AC$, since $angle CAE = 60° - 20° =40°$



    $angle AEC = frac{180° - 40°}{2} = 70°$



    so $x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Very nice +1......
      $endgroup$
      – greedoid
      8 hours ago











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    Construct an equilateral triangle such that its sides are equal to the base of the main triangle.
    enter image description here






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
      $endgroup$
      – greedoid
      8 hours ago












    • $begingroup$
      @greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
      $endgroup$
      – Seyed
      8 hours ago












    • $begingroup$
      Yes, that is correct if you want another upvote.
      $endgroup$
      – greedoid
      8 hours ago






    • 2




      $begingroup$
      @greedoid, Thanks for your advise.
      $endgroup$
      – Seyed
      8 hours ago
















    9












    $begingroup$

    Construct an equilateral triangle such that its sides are equal to the base of the main triangle.
    enter image description here






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
      $endgroup$
      – greedoid
      8 hours ago












    • $begingroup$
      @greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
      $endgroup$
      – Seyed
      8 hours ago












    • $begingroup$
      Yes, that is correct if you want another upvote.
      $endgroup$
      – greedoid
      8 hours ago






    • 2




      $begingroup$
      @greedoid, Thanks for your advise.
      $endgroup$
      – Seyed
      8 hours ago














    9












    9








    9





    $begingroup$

    Construct an equilateral triangle such that its sides are equal to the base of the main triangle.
    enter image description here






    share|cite|improve this answer











    $endgroup$



    Construct an equilateral triangle such that its sides are equal to the base of the main triangle.
    enter image description here







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 7 hours ago









    Acccumulation

    7,0372619




    7,0372619










    answered 9 hours ago









    SeyedSeyed

    7,01341424




    7,01341424












    • $begingroup$
      Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
      $endgroup$
      – greedoid
      8 hours ago












    • $begingroup$
      @greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
      $endgroup$
      – Seyed
      8 hours ago












    • $begingroup$
      Yes, that is correct if you want another upvote.
      $endgroup$
      – greedoid
      8 hours ago






    • 2




      $begingroup$
      @greedoid, Thanks for your advise.
      $endgroup$
      – Seyed
      8 hours ago


















    • $begingroup$
      Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
      $endgroup$
      – greedoid
      8 hours ago












    • $begingroup$
      @greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
      $endgroup$
      – Seyed
      8 hours ago












    • $begingroup$
      Yes, that is correct if you want another upvote.
      $endgroup$
      – greedoid
      8 hours ago






    • 2




      $begingroup$
      @greedoid, Thanks for your advise.
      $endgroup$
      – Seyed
      8 hours ago
















    $begingroup$
    Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
    $endgroup$
    – greedoid
    8 hours ago






    $begingroup$
    Well, both your's solution are beatifull and I can not upvote both if you write them in the same answer.
    $endgroup$
    – greedoid
    8 hours ago














    $begingroup$
    @greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
    $endgroup$
    – Seyed
    8 hours ago






    $begingroup$
    @greedoid, you mean I have to post them as two different answers? I am not really familiar with the voting rules.
    $endgroup$
    – Seyed
    8 hours ago














    $begingroup$
    Yes, that is correct if you want another upvote.
    $endgroup$
    – greedoid
    8 hours ago




    $begingroup$
    Yes, that is correct if you want another upvote.
    $endgroup$
    – greedoid
    8 hours ago




    2




    2




    $begingroup$
    @greedoid, Thanks for your advise.
    $endgroup$
    – Seyed
    8 hours ago




    $begingroup$
    @greedoid, Thanks for your advise.
    $endgroup$
    – Seyed
    8 hours ago











    6












    $begingroup$

    And my second solution is as follow:
    enter image description here






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      And my second solution is as follow:
      enter image description here






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        And my second solution is as follow:
        enter image description here






        share|cite|improve this answer









        $endgroup$



        And my second solution is as follow:
        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 8 hours ago









        SeyedSeyed

        7,01341424




        7,01341424























            4












            $begingroup$

            Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.



            Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.



            Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$



            Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
            $$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
            But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$



            Thus, $$B'C'=B'K=AD=BC,$$ which says that
            $$Bequiv B'$$ and $$Cequiv C'.$$
            Id est,
            $$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice as always +1
              $endgroup$
              – greedoid
              8 hours ago


















            4












            $begingroup$

            Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.



            Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.



            Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$



            Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
            $$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
            But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$



            Thus, $$B'C'=B'K=AD=BC,$$ which says that
            $$Bequiv B'$$ and $$Cequiv C'.$$
            Id est,
            $$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice as always +1
              $endgroup$
              – greedoid
              8 hours ago
















            4












            4








            4





            $begingroup$

            Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.



            Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.



            Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$



            Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
            $$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
            But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$



            Thus, $$B'C'=B'K=AD=BC,$$ which says that
            $$Bequiv B'$$ and $$Cequiv C'.$$
            Id est,
            $$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$






            share|cite|improve this answer











            $endgroup$



            Let in $Delta ABC$ we have $AB=AC$, $measuredangle A=20^{circ}$ and $measuredangle ADC=x$ as on your picture.



            Let $Min AB$ such that $AD=MD$ and $Kin DC$ such that $MK=AD$.



            Also, let $B'in MB$ such that $MB'=AD$ and $C'in KC$ such that $B'C'||BC.$



            Thus, $$measuredangle MKA=measuredangle MDK=2cdot20^{circ}=40^{circ}$$ and from here
            $$measuredangle B'MK=40^{circ}+20^{circ}=60^{circ},$$ which says $$B'K=MB'=AD=BC.$$
            But $$measuredangle B'KC'=60^{circ}+20^{circ}=80^{circ}=measuredangle BCA=measuredangle B'C'A.$$



            Thus, $$B'C'=B'K=AD=BC,$$ which says that
            $$Bequiv B'$$ and $$Cequiv C'.$$
            Id est,
            $$measuredangle BDC=10^{circ}+20^{circ}=30^{circ}.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 9 hours ago

























            answered 9 hours ago









            Michael RozenbergMichael Rozenberg

            106k1893198




            106k1893198












            • $begingroup$
              Nice as always +1
              $endgroup$
              – greedoid
              8 hours ago




















            • $begingroup$
              Nice as always +1
              $endgroup$
              – greedoid
              8 hours ago


















            $begingroup$
            Nice as always +1
            $endgroup$
            – greedoid
            8 hours ago






            $begingroup$
            Nice as always +1
            $endgroup$
            – greedoid
            8 hours ago













            4












            $begingroup$

            enter image description here



            construct triangle $Delta BCE$ congruent to $Delta ADB$.



            so $AB = BE$, $angle ABE = 80° - 20° = 60°$



            Thus triangle $Delta ABE$ is equilateral.



            $AB = AE = AC$, since $angle CAE = 60° - 20° =40°$



            $angle AEC = frac{180° - 40°}{2} = 70°$



            so $x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Very nice +1......
              $endgroup$
              – greedoid
              8 hours ago
















            4












            $begingroup$

            enter image description here



            construct triangle $Delta BCE$ congruent to $Delta ADB$.



            so $AB = BE$, $angle ABE = 80° - 20° = 60°$



            Thus triangle $Delta ABE$ is equilateral.



            $AB = AE = AC$, since $angle CAE = 60° - 20° =40°$



            $angle AEC = frac{180° - 40°}{2} = 70°$



            so $x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°$






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            4












            4








            4





            $begingroup$

            enter image description here



            construct triangle $Delta BCE$ congruent to $Delta ADB$.



            so $AB = BE$, $angle ABE = 80° - 20° = 60°$



            Thus triangle $Delta ABE$ is equilateral.



            $AB = AE = AC$, since $angle CAE = 60° - 20° =40°$



            $angle AEC = frac{180° - 40°}{2} = 70°$



            so $x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°$






            share|cite|improve this answer











            $endgroup$



            enter image description here



            construct triangle $Delta BCE$ congruent to $Delta ADB$.



            so $AB = BE$, $angle ABE = 80° - 20° = 60°$



            Thus triangle $Delta ABE$ is equilateral.



            $AB = AE = AC$, since $angle CAE = 60° - 20° =40°$



            $angle AEC = frac{180° - 40°}{2} = 70°$



            so $x = 20° + angle ABD = 20° + angle CBE = 20° + (70° - 60° ) = 30°$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 7 hours ago









            Dr. Mathva

            2,006324




            2,006324










            answered 8 hours ago









            qsmyqsmy

            31028




            31028












            • $begingroup$
              Very nice +1......
              $endgroup$
              – greedoid
              8 hours ago


















            • $begingroup$
              Very nice +1......
              $endgroup$
              – greedoid
              8 hours ago
















            $begingroup$
            Very nice +1......
            $endgroup$
            – greedoid
            8 hours ago




            $begingroup$
            Very nice +1......
            $endgroup$
            – greedoid
            8 hours ago










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