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Angle between two straight lines

  1. Mar 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the angle between the straight lines:
    ##(x^2+y^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(x \tan{\theta} - y \sin{\alpha})^2##

    2. Relevant equations
    [Not applicable]

    3. The attempt at a solution
    Dividing by ##x^2##,
    ## (1+(\frac{y}{x})^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(\tan{\theta} - \frac{y}{x} \sin{\alpha})^2 ##
    Let, ##\frac{y}{x} =m ##.
    ## (1+m^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(\tan{\theta} - m \sin{\alpha})^2 ##
    ##(\sin^2{\theta} \cos^2{\alpha}) m^2 + (2 \tan{\theta} \sin{\alpha}) m + (\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta}-\tan^2{\theta}) = 0 ##
    So the solutions of this equation indicate the slopes of the two straight lines. If the solutions are ##m_1## and ##m_2##, then the angle between the two straight lines will be ##\arctan{\frac{m_1-m_2}{1+m_1 m_2}}##;
    I came up with a messy equation, as I tried to calculate this. But the answer is very simple, just ##2\theta##. So, I think there is some clever technique to solve this problem.
    Any suggestion will be appreciated
     
  2. jcsd
  3. Mar 20, 2016 #2

    Mark44

    Staff: Mentor

    Instead of dividing by x^2, I think you might be better off by replacing x^2 + y^2 by r^2, and replacing x and y by r*cos(θ) and r*sin(θ), respectively. This would get the equation completely into polar form, after which it might be easier to simplify.
     
  4. Mar 20, 2016 #3
    Thanks for your reply. This also gives me same kind of equation.
     
  5. Mar 20, 2016 #4

    Mark44

    Staff: Mentor

    Doing what I suggested, I get a somewhat simpler equation in polar form.

    Is there more to this problem than you have posted? Is there an explanation of what ##\theta## and ##\alpha## represent?

    Also, is the equation you wrote exactly the same as given in the problem?
     
  6. Mar 20, 2016 #5
    No, there is no explanation about ##\theta## & ##\alpha##. And the equation I wrote is exactly the same as in the problem.
     
  7. Mar 20, 2016 #6

    Mark44

    Staff: Mentor

    Can you post an image of the problem? Your solution and my solution both seem to come out very messy, and I'm wondering if there is something we're not seeing.
     
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