A person standing on the ground observes
the angle of eleva tion of the top of a tower to be 30°. On
walking a distance a in a certain direction, he finds the elevation
of the top to be the same as before. He then walks
a distance $5a/3$ at right angles to his former direction, and
finds that the elevation of the top has doubled. The height
of the tower is

Let $PQ$ be the tower of height $h$and $A, B, C$
be the three positions of the person from initial to final then
(Fig. 27.48)

$\begin{array}{l}\mathrm{\angle }PAQ=\mathrm{\angle }PBQ={30}^{\circ }\text{ and }\mathrm{\angle }PCQ={60}^{\circ }\\ \Rightarrow AQ=BQ=h\cot {30}^{\circ }=h\sqrt{3}\\ \text{ and }CQ=h\cot {60}^{\circ }\models h/\sqrt{3}\end{array}$

Let $QM$ be perpendicular to$AB$ and$CL$ be perpendicular
to $QM$

Now $AB=a\text{ and }BC=5a/3\text{ and }\mathrm{\angle }ABC={90}^{\circ }$

$\Rightarrow ML=BC=5a/3\text{ and }CL=BM=a/2$

From right angled triangle $BMQ$,

$Q{M}^2=B{Q}^2-B{M}^2=3h^2-(a/2{)}^2=3h^2-\left(a^2/4\right)$

and from right angled triangle $CLQ.$

$Q{L}^2=Q{C}^2-C{L}^2=\frac{h^2}{3}-\frac{a^2}{4}$

Since  $QM=QL+LM$

$\begin{array}{l}\Rightarrow 3h^2-\frac{a^2}{4}=\frac{h^2}{3}-\frac{a^2}{4}+\frac{25a^2}{9}+\frac{10a}{3}\sqrt{\frac{h^2}{3}-\frac{a^2}{4}}\\ \Rightarrow \frac{8h^2}{3}-\frac{25a^2}{9}=\frac{10a}{3}\sqrt{\frac{h^2}{3}-\frac{a^2}{4}}\\ \Rightarrow 64h^4-\frac{400a^2{h}^2}{3}+\frac{625a^4}{9}=\frac{100a^2{h}^2}{3}-25a^4\\ \Rightarrow 576h^4-1500a^2{h}^2+850a^4=0\\ \Rightarrow 288h^4-750a^2{h}^2+425a^4=0\\ \Rightarrow \left(48h^2-85a^2\right)\left(6h^2-5a^2\right)=0\\ \Rightarrow h=\sqrt{85/48}a\text{ or }\sqrt{5/6}\cdot a\end{array}$