A vertical tower stands at an arbitrary point inside a rectangular...

Let the height of the vertical tower be $h and its base be at point$P$in the plane of the rectangle$ABCD$.<br><br>The horizontal distances from$P$to the corners$A$,$B$, and$C$can be found using the angles of elevation:<br>$PA = h \cot 30^\circ = h\sqrt{3}$<br>$PB = h \cot 45^\circ = h$<br>$PC = h \cot 60^\circ = \dfrac{h}{\sqrt{3}}$<br><br>For any point$P$in the plane of a rectangle$ABCD$, the distances to the vertices satisfy the relation:<br>$PA^2 + PC^2 = PB^2 + PD^2$<br><br>Substituting the values of$PA$,$PB$, and$PC$:<br>$(h\sqrt{3})^2 + \left(\dfrac{h}{\sqrt{3}}\right)^2 = h^2 + PD^2$<br>$3h^2 + \dfrac{h^2}{3} = h^2 + PD^2$<br>$PD^2 = 3h^2 + \dfrac{h^2}{3} - h^2 = \dfrac{7h^2}{3}$<br>$PD = h\sqrt{\dfrac{7}{3}}$<br><br>Let the angle of elevation of the top of the tower from$D$be$\delta$. Then,<br>$\tan \delta = \dfrac{h}{PD} = \dfrac{h}{h\sqrt{\dfrac{7}{3}}} = \sqrt{\dfrac{3}{7}}$<br><br>$\Rightarrow \delta = \tan^{-1}\left(\sqrt{\dfrac{3}{7}}\right)$<br><br>Answer:$\tan^{-1}\left(\sqrt{\dfrac{3}{7}}\right)