$A$ vertical tower CP subtends the same
angle$\theta$ at point $B$on the horizontal plane through $C$, the
foot of the tower, and at point $A$in the vertical plane. If the
triangle $ABC$ is equilateral with length of each side equal to
4 m, the height of the tower is 

Since$\angle CBP = \angle CAP = \theta$ (angles in the same
segment are equal) $PCBA$ is a cyclic quadrilateral. (Fig.
27.23)

$\Rightarrow \mathrm{\angle }BAP+\mathrm{\angle }BCP={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BAP={90}^{\circ } \left[\because \mathrm{\angle }BCP={90}^{\circ }\right]$

$\begin{array}{l}\Rightarrow \mathrm{\angle }PBC=\theta =\mathrm{\angle }BAP-\mathrm{\angle }BAC\\ ={90}^{\circ }-{60}^{\circ }={30}^{\circ }\end{array}$

If $h$is the height of the tower $CP,$ then $\frac{CP}{CB}=\tan \theta =$

$\tan A$

$\Rightarrow \frac{h}{4}=\frac{1}{\sqrt{3}}\Rightarrow h=\frac{4}{\sqrt{3}}\mathrm{m}$.