In $Delta ABC$ if $(sqrt{3}-1)a=2b$, $A=3B$, then find $C$












2












$begingroup$



In $Delta ABC$ if $(sqrt{3}-1)a=2b$, $A=3B$, then find $C$




My Attempt
$$
b=frac{sqrt{3}-1}{2}aquad& quad frac{A-B}{2}=Bquad&quadfrac{A+B}{2}=2B\frac{a-b}{a+b}=frac{tanfrac{A-B}{2}}{tanfrac{A+B}{2}}implies frac{3-sqrt{3}}{sqrt{3}+1}=frac{tan B}{tan 2B}=frac{1-tan^2B}{2}
$$










share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    In $Delta ABC$ if $(sqrt{3}-1)a=2b$, $A=3B$, then find $C$




    My Attempt
    $$
    b=frac{sqrt{3}-1}{2}aquad& quad frac{A-B}{2}=Bquad&quadfrac{A+B}{2}=2B\frac{a-b}{a+b}=frac{tanfrac{A-B}{2}}{tanfrac{A+B}{2}}implies frac{3-sqrt{3}}{sqrt{3}+1}=frac{tan B}{tan 2B}=frac{1-tan^2B}{2}
    $$










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      In $Delta ABC$ if $(sqrt{3}-1)a=2b$, $A=3B$, then find $C$




      My Attempt
      $$
      b=frac{sqrt{3}-1}{2}aquad& quad frac{A-B}{2}=Bquad&quadfrac{A+B}{2}=2B\frac{a-b}{a+b}=frac{tanfrac{A-B}{2}}{tanfrac{A+B}{2}}implies frac{3-sqrt{3}}{sqrt{3}+1}=frac{tan B}{tan 2B}=frac{1-tan^2B}{2}
      $$










      share|cite|improve this question











      $endgroup$





      In $Delta ABC$ if $(sqrt{3}-1)a=2b$, $A=3B$, then find $C$




      My Attempt
      $$
      b=frac{sqrt{3}-1}{2}aquad& quad frac{A-B}{2}=Bquad&quadfrac{A+B}{2}=2B\frac{a-b}{a+b}=frac{tanfrac{A-B}{2}}{tanfrac{A+B}{2}}implies frac{3-sqrt{3}}{sqrt{3}+1}=frac{tan B}{tan 2B}=frac{1-tan^2B}{2}
      $$







      geometry trigonometry euclidean-geometry triangle






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 21:49









      Michael Rozenberg

      101k1591193




      101k1591193










      asked Jan 4 at 21:21









      ss1729ss1729

      1,9261723




      1,9261723






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          In the standard notation by law of sines we obtain:
          $$frac{sinbeta}{sin3beta}=frac{sqrt3-1}{2}$$ or
          $$frac{1}{3-4sin^2beta}=frac{sqrt3-1}{2}$$ or
          $$3-4sin^2beta=sqrt3+1$$ or
          $$8sin^2beta=(sqrt3-1)^2$$ or
          $$sinbeta=frac{sqrt3-1}{2sqrt2},$$ which gives $$beta=15^{circ}.$$
          Can you end it now?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I would've gone with $ 2 - 4sin^2 B = sqrt{3} implies 2cos(2B) = sqrt{3} implies 2B = 30^circ $, since $sin 15^circ$ isn't really well-known
            $endgroup$
            – Dylan
            Jan 5 at 11:26












          • $begingroup$
            @Dylan Or maybe $2beta=330^{circ}$, which is $beta=165^{circ},$ which is impossible. I think everyone, which solved during his life one ore maybe two problems, knows what is it $sin15^{circ}.$
            $endgroup$
            – Michael Rozenberg
            Jan 5 at 12:40










          • $begingroup$
            Well yes, there would be 2 solutions, one of which will be ruled out since $3B < 180^circ$. I've never had to solve for $sin 15^circ$ so I don't know what you're talking about.
            $endgroup$
            – Dylan
            Jan 5 at 13:37












          • $begingroup$
            @Dylan There are also $sin18^{circ}$, $cos36^{circ}$, $tan15^{circ}$, $tan22.5^{circ}$ and more. It's better to know it.
            $endgroup$
            – Michael Rozenberg
            Jan 5 at 13:40












          • $begingroup$
            I suppose you're right, but I don't think it should be considered common knowledge.
            $endgroup$
            – Dylan
            Jan 5 at 13:46



















          1












          $begingroup$

          $$2sin B=(sqrt3-1)sin3B=(...)(sin B)(3-4sin^2B)$$



          As $sin B>0,$ $$sin^2B=dfrac{2-sqrt3}4$$



          $$cos2B=1-dfrac{2-sqrt3}2=cos30^circ$$



          $0<2B<360^circ,2B=360^circ npm30^circ$ for some integer $n$



          $implies2B=30^circ$






          share|cite|improve this answer









          $endgroup$













            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
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062125%2fin-delta-abc-if-sqrt3-1a-2b-a-3b-then-find-c%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            In the standard notation by law of sines we obtain:
            $$frac{sinbeta}{sin3beta}=frac{sqrt3-1}{2}$$ or
            $$frac{1}{3-4sin^2beta}=frac{sqrt3-1}{2}$$ or
            $$3-4sin^2beta=sqrt3+1$$ or
            $$8sin^2beta=(sqrt3-1)^2$$ or
            $$sinbeta=frac{sqrt3-1}{2sqrt2},$$ which gives $$beta=15^{circ}.$$
            Can you end it now?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I would've gone with $ 2 - 4sin^2 B = sqrt{3} implies 2cos(2B) = sqrt{3} implies 2B = 30^circ $, since $sin 15^circ$ isn't really well-known
              $endgroup$
              – Dylan
              Jan 5 at 11:26












            • $begingroup$
              @Dylan Or maybe $2beta=330^{circ}$, which is $beta=165^{circ},$ which is impossible. I think everyone, which solved during his life one ore maybe two problems, knows what is it $sin15^{circ}.$
              $endgroup$
              – Michael Rozenberg
              Jan 5 at 12:40










            • $begingroup$
              Well yes, there would be 2 solutions, one of which will be ruled out since $3B < 180^circ$. I've never had to solve for $sin 15^circ$ so I don't know what you're talking about.
              $endgroup$
              – Dylan
              Jan 5 at 13:37












            • $begingroup$
              @Dylan There are also $sin18^{circ}$, $cos36^{circ}$, $tan15^{circ}$, $tan22.5^{circ}$ and more. It's better to know it.
              $endgroup$
              – Michael Rozenberg
              Jan 5 at 13:40












            • $begingroup$
              I suppose you're right, but I don't think it should be considered common knowledge.
              $endgroup$
              – Dylan
              Jan 5 at 13:46
















            2












            $begingroup$

            In the standard notation by law of sines we obtain:
            $$frac{sinbeta}{sin3beta}=frac{sqrt3-1}{2}$$ or
            $$frac{1}{3-4sin^2beta}=frac{sqrt3-1}{2}$$ or
            $$3-4sin^2beta=sqrt3+1$$ or
            $$8sin^2beta=(sqrt3-1)^2$$ or
            $$sinbeta=frac{sqrt3-1}{2sqrt2},$$ which gives $$beta=15^{circ}.$$
            Can you end it now?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I would've gone with $ 2 - 4sin^2 B = sqrt{3} implies 2cos(2B) = sqrt{3} implies 2B = 30^circ $, since $sin 15^circ$ isn't really well-known
              $endgroup$
              – Dylan
              Jan 5 at 11:26












            • $begingroup$
              @Dylan Or maybe $2beta=330^{circ}$, which is $beta=165^{circ},$ which is impossible. I think everyone, which solved during his life one ore maybe two problems, knows what is it $sin15^{circ}.$
              $endgroup$
              – Michael Rozenberg
              Jan 5 at 12:40










            • $begingroup$
              Well yes, there would be 2 solutions, one of which will be ruled out since $3B < 180^circ$. I've never had to solve for $sin 15^circ$ so I don't know what you're talking about.
              $endgroup$
              – Dylan
              Jan 5 at 13:37












            • $begingroup$
              @Dylan There are also $sin18^{circ}$, $cos36^{circ}$, $tan15^{circ}$, $tan22.5^{circ}$ and more. It's better to know it.
              $endgroup$
              – Michael Rozenberg
              Jan 5 at 13:40












            • $begingroup$
              I suppose you're right, but I don't think it should be considered common knowledge.
              $endgroup$
              – Dylan
              Jan 5 at 13:46














            2












            2








            2





            $begingroup$

            In the standard notation by law of sines we obtain:
            $$frac{sinbeta}{sin3beta}=frac{sqrt3-1}{2}$$ or
            $$frac{1}{3-4sin^2beta}=frac{sqrt3-1}{2}$$ or
            $$3-4sin^2beta=sqrt3+1$$ or
            $$8sin^2beta=(sqrt3-1)^2$$ or
            $$sinbeta=frac{sqrt3-1}{2sqrt2},$$ which gives $$beta=15^{circ}.$$
            Can you end it now?






            share|cite|improve this answer









            $endgroup$



            In the standard notation by law of sines we obtain:
            $$frac{sinbeta}{sin3beta}=frac{sqrt3-1}{2}$$ or
            $$frac{1}{3-4sin^2beta}=frac{sqrt3-1}{2}$$ or
            $$3-4sin^2beta=sqrt3+1$$ or
            $$8sin^2beta=(sqrt3-1)^2$$ or
            $$sinbeta=frac{sqrt3-1}{2sqrt2},$$ which gives $$beta=15^{circ}.$$
            Can you end it now?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 21:28









            Michael RozenbergMichael Rozenberg

            101k1591193




            101k1591193












            • $begingroup$
              I would've gone with $ 2 - 4sin^2 B = sqrt{3} implies 2cos(2B) = sqrt{3} implies 2B = 30^circ $, since $sin 15^circ$ isn't really well-known
              $endgroup$
              – Dylan
              Jan 5 at 11:26












            • $begingroup$
              @Dylan Or maybe $2beta=330^{circ}$, which is $beta=165^{circ},$ which is impossible. I think everyone, which solved during his life one ore maybe two problems, knows what is it $sin15^{circ}.$
              $endgroup$
              – Michael Rozenberg
              Jan 5 at 12:40










            • $begingroup$
              Well yes, there would be 2 solutions, one of which will be ruled out since $3B < 180^circ$. I've never had to solve for $sin 15^circ$ so I don't know what you're talking about.
              $endgroup$
              – Dylan
              Jan 5 at 13:37












            • $begingroup$
              @Dylan There are also $sin18^{circ}$, $cos36^{circ}$, $tan15^{circ}$, $tan22.5^{circ}$ and more. It's better to know it.
              $endgroup$
              – Michael Rozenberg
              Jan 5 at 13:40












            • $begingroup$
              I suppose you're right, but I don't think it should be considered common knowledge.
              $endgroup$
              – Dylan
              Jan 5 at 13:46


















            • $begingroup$
              I would've gone with $ 2 - 4sin^2 B = sqrt{3} implies 2cos(2B) = sqrt{3} implies 2B = 30^circ $, since $sin 15^circ$ isn't really well-known
              $endgroup$
              – Dylan
              Jan 5 at 11:26












            • $begingroup$
              @Dylan Or maybe $2beta=330^{circ}$, which is $beta=165^{circ},$ which is impossible. I think everyone, which solved during his life one ore maybe two problems, knows what is it $sin15^{circ}.$
              $endgroup$
              – Michael Rozenberg
              Jan 5 at 12:40










            • $begingroup$
              Well yes, there would be 2 solutions, one of which will be ruled out since $3B < 180^circ$. I've never had to solve for $sin 15^circ$ so I don't know what you're talking about.
              $endgroup$
              – Dylan
              Jan 5 at 13:37












            • $begingroup$
              @Dylan There are also $sin18^{circ}$, $cos36^{circ}$, $tan15^{circ}$, $tan22.5^{circ}$ and more. It's better to know it.
              $endgroup$
              – Michael Rozenberg
              Jan 5 at 13:40












            • $begingroup$
              I suppose you're right, but I don't think it should be considered common knowledge.
              $endgroup$
              – Dylan
              Jan 5 at 13:46
















            $begingroup$
            I would've gone with $ 2 - 4sin^2 B = sqrt{3} implies 2cos(2B) = sqrt{3} implies 2B = 30^circ $, since $sin 15^circ$ isn't really well-known
            $endgroup$
            – Dylan
            Jan 5 at 11:26






            $begingroup$
            I would've gone with $ 2 - 4sin^2 B = sqrt{3} implies 2cos(2B) = sqrt{3} implies 2B = 30^circ $, since $sin 15^circ$ isn't really well-known
            $endgroup$
            – Dylan
            Jan 5 at 11:26














            $begingroup$
            @Dylan Or maybe $2beta=330^{circ}$, which is $beta=165^{circ},$ which is impossible. I think everyone, which solved during his life one ore maybe two problems, knows what is it $sin15^{circ}.$
            $endgroup$
            – Michael Rozenberg
            Jan 5 at 12:40




            $begingroup$
            @Dylan Or maybe $2beta=330^{circ}$, which is $beta=165^{circ},$ which is impossible. I think everyone, which solved during his life one ore maybe two problems, knows what is it $sin15^{circ}.$
            $endgroup$
            – Michael Rozenberg
            Jan 5 at 12:40












            $begingroup$
            Well yes, there would be 2 solutions, one of which will be ruled out since $3B < 180^circ$. I've never had to solve for $sin 15^circ$ so I don't know what you're talking about.
            $endgroup$
            – Dylan
            Jan 5 at 13:37






            $begingroup$
            Well yes, there would be 2 solutions, one of which will be ruled out since $3B < 180^circ$. I've never had to solve for $sin 15^circ$ so I don't know what you're talking about.
            $endgroup$
            – Dylan
            Jan 5 at 13:37














            $begingroup$
            @Dylan There are also $sin18^{circ}$, $cos36^{circ}$, $tan15^{circ}$, $tan22.5^{circ}$ and more. It's better to know it.
            $endgroup$
            – Michael Rozenberg
            Jan 5 at 13:40






            $begingroup$
            @Dylan There are also $sin18^{circ}$, $cos36^{circ}$, $tan15^{circ}$, $tan22.5^{circ}$ and more. It's better to know it.
            $endgroup$
            – Michael Rozenberg
            Jan 5 at 13:40














            $begingroup$
            I suppose you're right, but I don't think it should be considered common knowledge.
            $endgroup$
            – Dylan
            Jan 5 at 13:46




            $begingroup$
            I suppose you're right, but I don't think it should be considered common knowledge.
            $endgroup$
            – Dylan
            Jan 5 at 13:46











            1












            $begingroup$

            $$2sin B=(sqrt3-1)sin3B=(...)(sin B)(3-4sin^2B)$$



            As $sin B>0,$ $$sin^2B=dfrac{2-sqrt3}4$$



            $$cos2B=1-dfrac{2-sqrt3}2=cos30^circ$$



            $0<2B<360^circ,2B=360^circ npm30^circ$ for some integer $n$



            $implies2B=30^circ$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $$2sin B=(sqrt3-1)sin3B=(...)(sin B)(3-4sin^2B)$$



              As $sin B>0,$ $$sin^2B=dfrac{2-sqrt3}4$$



              $$cos2B=1-dfrac{2-sqrt3}2=cos30^circ$$



              $0<2B<360^circ,2B=360^circ npm30^circ$ for some integer $n$



              $implies2B=30^circ$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $$2sin B=(sqrt3-1)sin3B=(...)(sin B)(3-4sin^2B)$$



                As $sin B>0,$ $$sin^2B=dfrac{2-sqrt3}4$$



                $$cos2B=1-dfrac{2-sqrt3}2=cos30^circ$$



                $0<2B<360^circ,2B=360^circ npm30^circ$ for some integer $n$



                $implies2B=30^circ$






                share|cite|improve this answer









                $endgroup$



                $$2sin B=(sqrt3-1)sin3B=(...)(sin B)(3-4sin^2B)$$



                As $sin B>0,$ $$sin^2B=dfrac{2-sqrt3}4$$



                $$cos2B=1-dfrac{2-sqrt3}2=cos30^circ$$



                $0<2B<360^circ,2B=360^circ npm30^circ$ for some integer $n$



                $implies2B=30^circ$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 2:45









                lab bhattacharjeelab bhattacharjee

                225k15157275




                225k15157275






























                    draft saved

                    draft discarded




















































                    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%2f3062125%2fin-delta-abc-if-sqrt3-1a-2b-a-3b-then-find-c%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

                    Human spaceflight

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    張江高科駅