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Math Help - Snell's law using triangles,

  1. #1
    Junior Member
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    Snell's law using triangles,

    Hi,

    Given two right angle triangles sharing a common side AC.
    Call them ABC, and ADC.

    First triangle angle ACB is \frac{ \pi }{2}-\theta
    angle ABC is a right angle.

    Second triangle angle CAD is \frac{ \pi }{2}-\alpha
    angle ADC is a right angle.

    Why is the following statement true.

    <br /> <br />
\frac{BC}{cos(\frac{\pi}{2}-\theta)} = \frac{AD}{cos(\frac{\pi}{2}-\alpha)}<br /> <br />

    I don't see how this can hold when \theta \neq \alpha

    I tried using pythagoras, but that doesnt' work.
    Perhaps I am missing something about the angle of refraction.
    Last edited by craigmain; July 26th 2009 at 05:10 AM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have \widehat{BAC}=\theta, \ \widehat{ACD}=\alpha

    In triangle ABC use the sine law.

    \frac{BC}{\sin\theta}=\frac{AC}{\sin B}

    But \sin\theta=\cos\left(\frac{\pi}{2}-\theta\right) and \sin B=1

    Then \frac{BC}{\cos\left(\frac{\pi}{2}-\theta\right)}=AC

    Similarly for the triangle ADC:

    \frac{AD}{\cos\left(\frac{\pi}{2}-\alpha\right)}=AC
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