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Math Help - Tower Question

  1. #1
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    Tower Question

    The 3d triangles are getting me!!





    For the first part by considering appropriate isosceles triangles and right-angled triangles I got h = a \sin \alpha \sec \theta is that correct?

    I'm also lost on the next part as well.

    Bobak
    Attached Thumbnails Attached Thumbnails Tower Question-picture-14.png   Tower Question-picture-15.png  
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  2. #2
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    Opalg's Avatar
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    Here's a plan of the view on the ground:

    \setlength{\unitlength}{3.5mm}<br />
\begin{picture}(30,18)<br />
\put(2,0){\line(0,1){15}}<br />
\put(2,0){\line(5,3){19.5}}<br />
\put(2,15){\line(5,-3){12.6}}<br />
\put(2,15){\line(6,-1){19.5}}<br />
\put(2,15){\line(3,-5){6.6}}<br />
\put(8,4.8){\line(5,3){1.1}}<br />
\put(9.5,4.5){\line(-3,5){0.62}}<br />
\put(0.5,0){$O$}<br />
\put(0.5,14.5){$T'$}<br />
\multiput(6,1.4)(6,3.6){3}{$a$}<br />
\put(2.5,1){$\theta$}<br />
\put(0.5,7){$x$}<br />
\put(14.5,6){$P$}<br />
\put(21.5,10.5){$Q$}<br />
\end{picture}

    Let OT' = x (T' denotes the base of the tower, so I'm using x to denote the horizontal distance from O to the base of the tower, not the inclined distance to the top of the tower). Then h/x = \tan\alpha. From the picture, a/x = \cos\theta. Therefore \frac ah = \frac{a/x}{h/x} = \frac{\cos\theta}{\tan\alpha}, and so h = a\tan\alpha\sec\theta.

    Next, \frac h{QT'} = \tan\beta, so QT'\,^2 = h^2\cot^2\beta = a^2\tan^2\alpha\cot^2\beta\sec^2\theta. But by Pythagoras (see the above diagram), QT'\,^2 = a^2(4+\tan^2\theta). Thus \tan^2\alpha\cot^2\beta\sec^2\theta = 4+\tan^2\theta, and I guess you can take it from there to get the formula for tan^2(θ). Once you have that, the remaining two parts of the question are easy.
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  3. #3
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    Thank you very much Opalg I understand the error in taking the inclined lengths.

    just to check

    c) use tan^2 \theta  > 0

    d) \sqrt 2 : \sqrt 5

    Bobak

    Off topic: how did you produce the latex drawing ?
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  4. #4
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    Krizalid's Avatar
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    Quote Originally Posted by bobak View Post

    Off topic: how did you produce the latex drawing ?
    They're called PStricks, and it's pretty complicated to dominate it but not impossible.

    See also, Asymptote here.
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  5. #5
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    Opalg's Avatar
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    Quote Originally Posted by bobak View Post
    Off topic: how did you produce the latex drawing ?
    I used the LaTeX picture environment, as described in Leslie Lamport's LaTeX manual. The code for that picture is

    [tex]\setlength{\unitlength}{3.5mm}
    \begin{picture}(30,18)
    \put(2,0){\line(0,1){15}}
    \put(2,0){\line(5,3){19.5}}
    \put(2,15){\line(5,-3){12.6}}
    \put(2,15){\line(6,-1){19.5}}
    \put(2,15){\line(3,-5){6.6}}
    \put(8,4.8){\line(5,3){1.1}}
    \put(9.5,4.5){\line(-3,5){0.62}}
    \put(0.5,0){$O$}
    \put(0.5,14.5){$T'$}
    \multiput(6,1.4)(6,3.6){3}{$a$}
    \put(2.5,1){$\theta$}
    \put(0.5,7){$x$}
    \put(14.5,6){$P$}
    \put(21.5,10.5){$Q$}
    \end{picture}[/tex]
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