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Tan(A+B) problem

  1. Jan 31, 2010 #1
    1. If sinA=7/25 and sinB=5/13, where A is acute and B is obtuse, find the exact value of tan(A+B)



    2. Tan(A+B)=tanA+tanB/1-tanAtanB



    3. TanA=7/24
    TanB=5/-12
    Tan(A+B)=7/24+5/-12/1-7/24x5/-12=-36/323 or -0.1 (1d.p.)


    Could somebody please check this for me, please?
     
  2. jcsd
  3. Jan 31, 2010 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi lemon! :smile:

    Very good :approve:, except :rolleyes:

    read the question … it asks for the exact value, which I assume means leave it as a fraction (exactly as the original data were given). :wink:
     
  4. Jan 31, 2010 #3

    Mark44

    Staff: Mentor

    Re: Tan(A+B)

    Removed the extra bold tags...
    Please use parentheses. You have everything jammed together, so it's difficult to tell what's in the numerator and what's in the denominator. This should be written as

    tan(A + B) = (tan A + tan B)/(1 - tan A * tan B)
    Again, please use parentheses. It would be clearer as

    tan(A + B) = (7/24 - 5/12)/(1 - (7/24)(-5/12))
    Your answer of -36/323 [itex]\approx[/itex] -0.111455 [itex]\approx[/itex] -0.1 is correct.
     
  5. Jan 31, 2010 #4
    Re: Tan(A+B)

    Understood. Thanks to you both
     
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