$A$tower $PQ$ subtends an angle $\alpha$ at a point
$A$on the same level as the foot $Q$of the tower. It also subtends
the same angle $\alpha$ at a point $B$where $AB$ subtends the
angle a with$AP$ then 

Since $\angle PAQ = \angle PBQ = a$ (angles in the
same segment are equal)$PQAB$ is a cyclic quadrilateral
(Fig. 27.49)

Also since $\mathrm{\angle }PAB=\alpha$ given,$\mathrm{\angle }PQB=\alpha$ (angles in the
same segment.)

$\Rightarrow$Now $\mathrm{\angle }PBQ=\alpha =\mid \mathrm{\angle }PQB$

$\Rightarrow \mathrm{\angle }ABQ=\mathrm{\angle }AQB={90}^{\circ }-\alpha$

$\Rightarrow ABQ$ is an isosceles triangle.

$\Rightarrow AB=AQ,$ and 

$\begin{array}{l}\frac{AB}{\sin \left({90}^{\circ }-\alpha \right)}=\frac{BQ}{\sin 2\alpha }\Rightarrow 2AB\sin \alpha =BQ.\\ \Rightarrow \frac{AB}{BQ}=\frac{1}{2\sin \alpha }=\frac{1}{2}\mathrm{cosec}\alpha .\end{array}$