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mramahmedmram
mramahmedmram asked in Science & MathematicsMathematics · 4 years ago

triangle ABC,m(<A)=60 PROVE 3/(a + b +c) = 1/a+b + 1/ a+c?

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  • Philip
    Lv 6
    4 years ago

    In triangle ABC, angle A = 60 deg. Now a^2 = b^2 + c^2 - 2bc cos(A). Now

    cos(A) = cos(60deg) = (1/2). Then a^2 = b^2 + c^2 - bc...[1].

    RTP: 3/(a+b+c) =1/(a+b) +1/(a+c), ie., RTP: 3(a+b)(a+c) = (a+b+c)(2a+b+c), ie., RTP: 3(a+b)(a+c) - (a+b+c)(2a+b+c) = 0...[2].

    LS[2] = 3[a^2 +ac +ab +bc] - [2a^2 +ab +ac + 2ab +b^2 +bc +2ac +bc+c^2

    = 3a^2 +3ac +3ab +3bc -2a^2 -ab -ac -2ab -b^2 -bc -2ac -bc -c^2

    = a^2 -(b^2+c^2) + bc. Now LS[1]-RS[1] ---> a^2 -(b^2+c^2) + bc = 0.

    Therefore LS[2] = 0 = RS[2] & QED.

  • Indica
    Lv 7
    4 years ago

    By Cosine Rule a² = b²+c²−2bccos(60) = b²+c²−bc

    b²+c²−bc=a²

    b²+ab+c²+ac = a²+ab+ac+bc

    b(a+b)+c(a+c) = (a+c)(a+b)

    b/(a+c)+c/(a+b) = 1

    b/(a+c)+1+c/(a+b)+1 = 3

    (b+a+c)/(a+c) + (c+a+b)/(a+b) = 3

    1/(a+c)+1/(a+b) = 3/(a+b+c)

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