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    trigo help

    prove that if in triangle ABC tan(A/2)=5/6 tan(B/2)=2/5 then a,b,c will be in AP
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    Senior Member BAdhi's Avatar
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    what's "AP"
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    Quote Originally Posted by prasum View Post
    prove that if in triangle ABC tan(A/2)=5/6 tan(B/2)=2/5 then a,b,c will be in AP
    I see that you have over 100 other postings. So you should understand that this is not a homework service nor is it a tutorial service.
    PLease either post some of your own work on this problem or explain what you do not understand about the question.
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    i have found tan A and Tan B through tan2A formula after that i have found out Tan C by Tan A+TanB+TanC=TanATanBTanC brcause A+b+c=180 degrees after that how should i find a,b,c and prove they are in arithmetic progression
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  5. #5
    Senior Member BAdhi's Avatar
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    wouldn't sinA=\frac{2tan(A/2)}{1+tan^2(A/2)} and \frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)} work?
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  6. #6
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    Quote Originally Posted by prasum View Post
    prove that if in triangle ABC, tan(A/2)=5/6, tan(B/2)=2/5, then a,b,c will be in AP
    \displaystyle\ tan\left[\frac{A}{2}\right]=\frac{5}{6}\Rightarrow\ sinA=\frac{2tan\left[\frac{A}{2}\right]}{1+tan^2\left[\frac{A}{2}\right]}=\frac{\left[\frac{5}{3}\right]}{\left[\frac{61}{36}\right]}=\frac{60}{61}

    \displaystyle\ tan\left[\frac{B}{2}\right]=\frac{2}{5}\Rightarrow\ sinB=\frac{\left[\frac{4}{5}\right]}{\left[\frac{29}{25}\right]}=\frac{20}{29}


    sinC=sin[180^o-(A+B)]=sin(A+B)=sinAcosB+cosAsinB


    We obtain the cosines from Pythagoras' Theorem.

    x^2+60^2=61^2\Rightarrow\ x^2=3721-3600=121\Rightarrow\ x=11

    \displaystyle\Rightarrow\ cosA=\frac{11}{61}

    y^2+20^2=29^2\Rightarrow\ y^2=29^2-20^2=441\Rightarrow\ y=21

    \displaystyle\Rightarrow\ cosB=\frac{21}{29}


    \displaystyle\ sinC=\frac{60}{61}\;\frac{21}{29}+\frac{11}{61}\;\  frac{20}{29}=\frac{1260+220}{1769}=\frac{1480}{176  9}


    From the Law of Sines

    \displaystyle\frac{a}{sinA}=\frac{b}{sinB}=\frac{c  }{sinC}

    \Rightarrow\ a:b:c=sinA:sinB:sinC

    If a, b, and c are in Arithmetic progression,
    then the sum of the 3 sides is 3 times the "middle side".

    p+(p+d)+(p+2d)=3p+3d=3(p+d)

    Hence, the sum of the 3 sines must be 3 times the sum of the "middle sine".

    sinA+sinB+sinC=2.50989259469

    3sinC=2.50989259469

    Therefore

    a, c, b are in Arithmetic Progression.
    b, c, a are in Arithmetic Progression.
    Last edited by Archie Meade; December 30th 2010 at 07:32 AM. Reason: typo
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