The 3d triangles are getting me!!
For the first part by considering appropriate isosceles triangles and right-angled triangles I got is that correct?
I'm also lost on the next part as well.
Bobak
Here's a plan of the view on the ground:
Let (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 . From the picture, . Therefore , and so .
Next, , so . But by Pythagoras (see the above diagram), Thus , 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.
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]