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Math Help - Trig & Pythagoras Triples

  1. #1
    Newbie Tom G's Avatar
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    Dec 2006
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    Trig & Pythagoras Triples

    Hi, got this question and I've got the answer=9/19, but the answer given in the book is 63/16.

    I am using the identity Tan(A+B)=(TanA+TanB)/(1-[TanATanB]). The question is as follows;

    Given that A and B are acute angles such that sinA=5/13 and cosB=3/5, find the exact value of Tan(A+B).

    I've got as far as the identity (shown above) and have TanA=5/12 and TanB=4/3, using pythagoras triples. However when I put this into the identity I get completly confused. Can someone check that my values for TanA and TanB are correct and show me how to complete the question. Thanks in advance everyone.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    Yes, you're correct. tanA = 5/12 and tanB = 4/3.

    You got lost in your simplifying the expression with all those fractions.

    tan(A+B) = [5/12 +4/3] /[1 -(5/12)(4/3)]
    = [(5 +16)/12] /[1 -20/36]
    = [21/12] /[1 -5/9]
    = [21/12] /[4/9]
    = [21/12]*[9/4]
    = [21*9] /[12*4]
    = [21*3] /[4*4]
    = 63/16
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