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












2















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




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






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    2















    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



























      2












      2








      2


      1






      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
















      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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      edited Jan 4 at 21:49









      Michael Rozenberg

      97.4k1589188




      97.4k1589188










      asked Jan 4 at 21:21









      ss1729ss1729

      1,8491723




      1,8491723






















          2 Answers
          2






          active

          oldest

          votes


















          2














          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





















          • 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
            – Dylan
            Jan 5 at 11:26












          • @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 $sin15^{circ}.$
            – Michael Rozenberg
            Jan 5 at 12:40










          • 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.
            – Dylan
            Jan 5 at 13:37












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












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



















          1














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



          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





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            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





















            • 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
              – Dylan
              Jan 5 at 11:26












            • @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 $sin15^{circ}.$
              – Michael Rozenberg
              Jan 5 at 12:40










            • 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.
              – Dylan
              Jan 5 at 13:37












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












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
















            2














            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





















            • 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
              – Dylan
              Jan 5 at 11:26












            • @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 $sin15^{circ}.$
              – Michael Rozenberg
              Jan 5 at 12:40










            • 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.
              – Dylan
              Jan 5 at 13:37












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












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














            2












            2








            2






            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












            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

            97.4k1589188




            97.4k1589188












            • 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
              – Dylan
              Jan 5 at 11:26












            • @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 $sin15^{circ}.$
              – Michael Rozenberg
              Jan 5 at 12:40










            • 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.
              – Dylan
              Jan 5 at 13:37












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












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


















            • 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
              – Dylan
              Jan 5 at 11:26












            • @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 $sin15^{circ}.$
              – Michael Rozenberg
              Jan 5 at 12:40










            • 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.
              – Dylan
              Jan 5 at 13:37












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












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
















            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
            – Dylan
            Jan 5 at 11:26






            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
            – Dylan
            Jan 5 at 11:26














            @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 $sin15^{circ}.$
            – Michael Rozenberg
            Jan 5 at 12:40




            @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 $sin15^{circ}.$
            – Michael Rozenberg
            Jan 5 at 12:40












            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.
            – Dylan
            Jan 5 at 13:37






            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.
            – Dylan
            Jan 5 at 13:37














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






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














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




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











            1














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



            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


























              1














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



              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
























                1












                1








                1






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



                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












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



                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

                224k15156274




                224k15156274






























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