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Problem involving trig functions

  1. Nov 19, 2007 #1
    [SOLVED] problem involving trig functions

    1. The problem statement, all variables and given/known data
    Your room has a window whose height is 1.5 meters. The bottom edge of the window is 10 cm above your eye level. (See figure) How far away from the window should you stand to get the best view? (“Best view” means the largest visual angle, i.e. angle between the lines of sight to the bottom and to the top of the window.)


    2. Relevant equations

    3. The attempt at a solution
    So the question's basically saying that I should maximize theta right? What I have so far is just
    tan(O+a) = 1.6/x
    a = arctan(0.1/x)
    [by the way, O=theta, i couldn't figure out how to insert the proper symbol]
    But even if I substitute a into the first equation, I still have 2 variables, and I can't figure out how to get only theta in an equation. Am I even on the right track?
  2. jcsd
  3. Nov 19, 2007 #2


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    Homework Helper

    have you tried using
    [tex]\tan (\theta+\alpha) = \frac{\tan \theta + \tan \alpha}{1-\tan \theta \,\tan \alpha}[/tex]

    and get an equation involving only x and tan theta, then to maximise, try determine global turning point (by looking at the derivaties)
  4. Nov 19, 2007 #3
    But when I take the derivative of tan (O+0.1/x) = 1.6/x, or i change it using that tan identity, I get an equation that involves theta, the derivative of theta, and x, and I'm not sure how to optimize that
  5. Nov 19, 2007 #4
    This is what i get when i take the derivative after using the tan identity:

    [tex]\frac{x^{2}(sec^{2}\theta\frac{d\theta}{dx} - \frac{0.1}{x^{2}})}{-0.1(xsec^{2}\theta\frac{d\theta}{dx} - tan\theta)} = \frac{-1.6}{x^{2}}[/tex]

    I still have both [tex]\theta[/tex] and [tex]x[/tex], so how do I optimize it?

    Edit: never mind, I got it. :)
    Last edited: Nov 20, 2007
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