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Thread: 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 $\displaystyle \frac{ \pi }{2}-\theta$
    angle ABC is a right angle.

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

    Why is the following statement true.

    $\displaystyle

    \frac{BC}{cos(\frac{\pi}{2}-\theta)} = \frac{AD}{cos(\frac{\pi}{2}-\alpha)}

    $

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

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

    In triangle ABC use the sine law.

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

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

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

    Similarly for the triangle ADC:

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