Question

A square tower stands upon a horizontal plane. From a point in this plane from which three of its upper corners are visible, their angular elevations are ${45}^o,{60}^o$ and ${45}^o$. If h is the height of the tower and $\alpha$ the breadth of one of its sides then show that $\frac{h}{a}=\frac{\sqrt{6}(\sqrt{5}+1)}{4}$

Answer 1

Let ABCD, A', B', C', D' represent the tower on square base ABCD. since the elevations of B' and D' at O are the same each equal to ${45}^o$, it follows that the position O of the observer must he on the diagonal CA of the base produced. The elevation of A' at O is ${60}^o$. Then from the figure it is clear that
$OA=h\cot {60}^o=\frac{h}{\sqrt{3}}$
and $OB=OD=h\cot {45}^o=h$.
Also, $\angle OAB={135}^o$
Hence from $\Delta OAB$, we have
$O{B}^2=O{A}^2+A{B}^2-2.OA.AB\cos {135}^o$
or $h^2=\frac{h^2}{3}+a^2-2\left(\frac{h^2}{3}\right).a\left(\frac{-1}{\sqrt{2}}\right)$
or $2h^2-\sqrt{6}ah-3a^2=0$
$\frac{h}{a}=\frac{\sqrt{6}(\sqrt{5}+1)}{4}$