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Angle Help!15 = arctan(2/x) - arctan (1/x)

  1. Jan 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Basically, solve for x
    15 = arctan(2/x) - arctan (1/x)

    2. Relevant equations
    tan (A-B) = (Tan A -Tan B) / (1+Tan A*Tan B)

    3. The attempt at a solution
    I really tried everything.
    My first step was to:
    Let y = arctan (2/x)
    Therefore, tan y = 2/x
    Similarly, u = 1/x
    Then, tan (y-u) = (Tan y -Tan u) / (1+Tan u*Tan y) = 15
    15 = (2/x-(1/x) / (1+(2/x^2)
    15 = (1/x) / (x^2 + 2 / x^2)
    15 = (x) / (x^2+2)
    15x^2 + 30 - x = 0
    Which has no real roots :(
    But, with guess and check, it's around 0.65
     
  2. jcsd
  3. Jan 22, 2011 #2

    rock.freak667

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    Homework Helper

    You need to post an attempt before we can help you. Start by taking the tangent of both sides.
     
  4. Jan 22, 2011 #3
    Yeah, sorry. I just misclicked the first time
     
  5. Jan 22, 2011 #4
    Alright, sorry guys to waste your time, but I believe I figured it out. Thanks for the hint of "tanning" both sides.
    Instead of 15, it's supposed to be tan 15.
    So that,
    x/(x^2+2) = tan 15
    x = 0.64
    x = 3.08 (approximately)

    Thanks for the help!
     
  6. Jan 22, 2011 #5

    Mentallic

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    Homework Helper

    If you rearrange that equation you'll get the quadratic

    [tex]x^2-\frac{1}{tan(15)}x+2=0[/tex]

    There are no real solutions to this quadratic.
     
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