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







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












$begingroup$


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











    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
    });


    }
    });






    Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3131581%2ffind-x-angle-in-triangle%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























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






            share|cite|improve this answer











            $endgroup$













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














            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










            Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.













            Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.












            Andriy Khrystyanovich is a new contributor. Be nice, and check out our Code of Conduct.
















            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%2f3131581%2ffind-x-angle-in-triangle%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

            If I really need a card on my start hand, how many mulligans make sense? [duplicate]

            Image
            5 This question already has an answer here: How do you calculate the likelihood of drawing certain cards in your opening hand? 3 answers There is a funny modern deck where you play Treasure Hunt, but you really need to have it on the starting hand. How many mulligans should I take, if my deck contains 60 cards and contains 4 copies of Treasure Hunt? magic-the-gathering probability share | improve this question asked Nov 23 '18 at 12:12 dan dan 152 4
            Read more

            Alcedinidae

            Image
            Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
            Read more

            Can an atomic nucleus contain both particles and antiparticles? [duplicate]

            Image
            9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
            Read more