# Tower Question

• Jun 21st 2008, 12:50 AM
bobak
Tower Question
The 3d triangles are getting me!!

http://www.mathhelpforum.com/math-he...picture-14.png

http://www.mathhelpforum.com/math-he...picture-15.png

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
• Jun 21st 2008, 03:32 AM
Opalg
Here's a plan of the view on the ground:

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

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.
• Jun 21st 2008, 04:06 AM
bobak
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 ?
• Jun 21st 2008, 08:44 AM
Krizalid
Quote:

Originally Posted by bobak

Off topic: how did you produce the latex drawing ?

They're called PStricks, and it's pretty complicated to dominate it but not impossible.

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