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Thread: Solving for x: A/sin(x)=B/sin(C-x)

  1. #1
    May 2011

    Question Solving for x: A/sin(x)=B/sin(C-x)


    I am trying to solve this equation: A/sin(x)=B/sin(C-x)
    for x, where 0<=C<=pi and A<B.

    It's been years since I last would have contemplated solving such an expression and quite frankly I am somewhat saddened that I have forgotten how. I once knew how to tackle this. Much of the maths I learnt studying engineering just never got used .

    It actually is a derivative of the law of sines equation relating to solving the torpedo fire control problem.

    x= angle of deflection
    A= target speed
    B= torpedo speed
    C= track angle

    I ultimately want to be able to use Excel to create and plot curves like these.

    Once I am on the right track I will be fine but I can't actually say what approach needs to be taken to deal with this problem. Need someone talking maths to me using maths language again. I can't even remember what this type of problem/expression is called.

    Thanks for helping out.



    PS: Not sure if this is in the right forum
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  2. #2
    e^(i*pi)'s Avatar
    Feb 2009
    West Midlands, England
    $\displaystyle A\sin(C-x) = B\sin(x)$

    $\displaystyle A(\sin(C) \cos(x) - \cos(C) \sin(x)) = B\sin(x)$

    $\displaystyle \sin(C) \cot(x) - \cos(C) = \dfrac{B}{A}$

    $\displaystyle \cot(x) = \csc(C) \left(\dfrac{B}{A} + \cos(C)\right)$

    $\displaystyle \tan(x) = \sin(C) \dfrac{1}{\left(\dfrac{B}{A} + \cos(C)\right)}$

    $\displaystyle x = \arctan \left( \dfrac{\sin(C)}{\left(\dfrac{B}{A} + \cos(C)\right)}\right)$
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  3. #3
    May 2011
    OK, as I kind of suspected, subbing in trignometric identities was the way to go. Actually, I would have preferred if you just said something like:

    Use a basic trig identity substitution (hint: sin(A-B)=sinA.cosB-cosA.sinB)
    Relax, it's not complex.

    ...and let me take it from there and salvage what dignity I have left.

    Still, thank you!
    Last edited by topsquark; May 1st 2011 at 10:29 AM.
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