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






geometry trigonometry euclidean-geometry triangle







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$endgroup$

















    2












    $begingroup$



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




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






    geometry trigonometry euclidean-geometry triangle







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






      geometry trigonometry euclidean-geometry triangle







      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




      geometry trigonometry euclidean-geometry triangle






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      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 4 at 21:49









      Michael Rozenberg

      106k1894198




      106k1894198










      asked Jan 4 at 21:21









      ss1729ss1729

      2,00111024




      2,00111024






















          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









          $endgroup$













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            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

            106k1894198




            106k1894198












            • $begingroup$
              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

                226k15157275




                226k15157275






























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