prove $a,b,c$ in A.P if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$

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In $Delta ABC$, if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.




My Attempt
$$
sin A=frac{2.5}{6}.frac{36}{61}=frac{60}{61}\
sin C=frac{2.2}{5}.frac{25}{29}=frac{20}{29}\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?






trigonometry triangle







share|cite|improve this question









$endgroup$

















    1












    $begingroup$



    In $Delta ABC$, if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.




    My Attempt
    $$
    sin A=frac{2.5}{6}.frac{36}{61}=frac{60}{61}\
    sin C=frac{2.2}{5}.frac{25}{29}=frac{20}{29}\
    $$
    it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?






    trigonometry triangle







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$



      In $Delta ABC$, if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.




      My Attempt
      $$
      sin A=frac{2.5}{6}.frac{36}{61}=frac{60}{61}\
      sin C=frac{2.2}{5}.frac{25}{29}=frac{20}{29}\
      $$
      it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?






      trigonometry triangle







      share|cite|improve this question









      $endgroup$





      In $Delta ABC$, if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.




      My Attempt
      $$
      sin A=frac{2.5}{6}.frac{36}{61}=frac{60}{61}\
      sin C=frac{2.2}{5}.frac{25}{29}=frac{20}{29}\
      $$
      it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?






      trigonometry triangle




      trigonometry triangle






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 16:37









      ss1729ss1729

      1,9211723




      1,9211723






















          4 Answers
          4






          active

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          0












          $begingroup$

          $$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$



          Use



          $$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            i was thinking of doing that straightaway .. but is there a better way other than this ?
            $endgroup$
            – ss1729
            Jan 4 at 16:50





















          0












          $begingroup$

          It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
          Can you finish now?






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Hint:



            Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,



            $$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              You can first deduce
              $$
              tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
              frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
              =frac{20}{37}
              $$
              Therefore
              $$
              sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
              $$
              Similarly,
              $$
              sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
              $$
              By the sine law,
              $$
              frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
              bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
              $$






              share|cite|improve this answer









              $endgroup$













                Your Answer





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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                0












                $begingroup$

                $$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$



                Use



                $$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  i was thinking of doing that straightaway .. but is there a better way other than this ?
                  $endgroup$
                  – ss1729
                  Jan 4 at 16:50


















                0












                $begingroup$

                $$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$



                Use



                $$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  i was thinking of doing that straightaway .. but is there a better way other than this ?
                  $endgroup$
                  – ss1729
                  Jan 4 at 16:50
















                0












                0








                0





                $begingroup$

                $$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$



                Use



                $$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$






                share|cite|improve this answer











                $endgroup$



                $$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$



                Use



                $$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 4 at 16:50

























                answered Jan 4 at 16:48









                lab bhattacharjeelab bhattacharjee

                225k15157275




                225k15157275












                • $begingroup$
                  i was thinking of doing that straightaway .. but is there a better way other than this ?
                  $endgroup$
                  – ss1729
                  Jan 4 at 16:50




















                • $begingroup$
                  i was thinking of doing that straightaway .. but is there a better way other than this ?
                  $endgroup$
                  – ss1729
                  Jan 4 at 16:50


















                $begingroup$
                i was thinking of doing that straightaway .. but is there a better way other than this ?
                $endgroup$
                – ss1729
                Jan 4 at 16:50






                $begingroup$
                i was thinking of doing that straightaway .. but is there a better way other than this ?
                $endgroup$
                – ss1729
                Jan 4 at 16:50













                0












                $begingroup$

                It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
                Can you finish now?






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
                  Can you finish now?






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
                    Can you finish now?






                    share|cite|improve this answer









                    $endgroup$



                    It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
                    Can you finish now?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 4 at 17:08









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    74.9k42865




                    74.9k42865























                        0












                        $begingroup$

                        Hint:



                        Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,



                        $$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Hint:



                          Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,



                          $$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Hint:



                            Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,



                            $$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$






                            share|cite|improve this answer









                            $endgroup$



                            Hint:



                            Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,



                            $$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 4 at 17:25









                            lab bhattacharjeelab bhattacharjee

                            225k15157275




                            225k15157275























                                0












                                $begingroup$

                                You can first deduce
                                $$
                                tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
                                frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
                                =frac{20}{37}
                                $$
                                Therefore
                                $$
                                sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
                                $$
                                Similarly,
                                $$
                                sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
                                $$
                                By the sine law,
                                $$
                                frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
                                bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
                                $$






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  You can first deduce
                                  $$
                                  tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
                                  frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
                                  =frac{20}{37}
                                  $$
                                  Therefore
                                  $$
                                  sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
                                  $$
                                  Similarly,
                                  $$
                                  sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
                                  $$
                                  By the sine law,
                                  $$
                                  frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
                                  bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    You can first deduce
                                    $$
                                    tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
                                    frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
                                    =frac{20}{37}
                                    $$
                                    Therefore
                                    $$
                                    sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
                                    $$
                                    Similarly,
                                    $$
                                    sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
                                    $$
                                    By the sine law,
                                    $$
                                    frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
                                    bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    You can first deduce
                                    $$
                                    tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
                                    frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
                                    =frac{20}{37}
                                    $$
                                    Therefore
                                    $$
                                    sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
                                    $$
                                    Similarly,
                                    $$
                                    sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
                                    $$
                                    By the sine law,
                                    $$
                                    frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
                                    bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
                                    $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 4 at 17:56









                                    egregegreg

                                    181k1485203




                                    181k1485203






























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