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Math Help - I can't figure out this triangle-based problem

  1. #1
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    I can't figure out this triangle-based problem

    I have included a drawing as an attachment.

    The following lengths are given: c, d, H and L

    The angle indicated with green is Theta. This is given too.

    The two red markings indicate 90 degree angles.

    I need to find length "a".

    I'm really stuck on this one. It's probably not that hard, but I could use some clues as it's only a small part of a bigger physics problem.
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  2. #2
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    Triangle trigonometry

    Hello Tovi

    Welcome to Math Help Forum!
    Quote Originally Posted by Tovi View Post
    I have included a drawing as an attachment.

    The following lengths are given: c, d, H and L

    The angle indicated with green is Theta. This is given too.

    The two red markings indicate 90 degree angles.

    I need to find length "a".

    I'm really stuck on this one. It's probably not that hard, but I could use some clues as it's only a small part of a bigger physics problem.
    I've added some detail to your drawing. See attachment.

    I assume that the line CG is perpendicular to AB. In which case \angle CDF = \angle CBG = \phi, say.

    Then \tan \theta = \frac{h}{c} and \tan\phi = \frac{h}{d}

    And CD = \frac{a}{\tan\theta}=\frac{ac}{h} = \frac{L-a}{\tan\phi}=\frac{d(L-a)}{h}

    \Rightarrow ac = d(L-a)=dL-da

    \Rightarrow a = \frac{dL}{c+d}

    Looks pretty simple when you see it, doesn't it?

    Grandad
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  3. #3
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    Nice explanation Grandad.

    I assume that the line CG is perpendicular to AB

    That simplifies things considerably, but it is solvable otherwise.
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  4. #4
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    Hello aidan
    Quote Originally Posted by aidan View Post
    Nice explanation Grandad.


    That simplifies things considerably, but it is solvable otherwise.
    In fact I was wrong in saying 'I assume...' because we can deduce that CG and AB are perpendicular, since \angle EDC = \angle CAG = \theta (given), and ED and AC are given perpendicular. So by angle sum of triangle (or indeed the quadrilateral) \angle AGD = 90^o.

    Grandad
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