$OAB$ is a triangle in the horizontal plane
through the foot $P$of the tower at the middle point of the
side $OB$ of the triangle. If$OA = 2 m, OB = 6 m, AB = 5 m$
and $\angle AOB$ is equal to the angle sub tended by the tower at
$A$then the height of the tower is 

Let$PQ$be the tower of height $h$at the middle
point $P$of the side $OB$ of the triangle$OAB,$ where (Fig.
27.46)

$OA=2,OB=6,AB=5\text{ and }\mathrm{\angle }AOB=\mathrm{\angle }PAQ=\alpha ,$

then $AP=h\cot \alpha ,OP=3$

From triangle$OAB,$$\cos \alpha =\frac{2^2+6^2-5^2}{2\times 2\times 6}=\frac{5}{8}$

and from $∆ OAP$

$\begin{array}{l}\cos \alpha =\frac{2^2+3^2-h^2{\cot }^2\alpha }{2\times 2\times 3}\\ \Rightarrow \frac{5}{8}=\frac{13-h^2\times \frac{25}{39}}{12}\\ \Rightarrow h^2=\frac{11\times 39}{25\times 2}\end{array}$