Two vertical poles 20 m and 80 m high stand apart on a horizontal plane. The height of the point of intersection of the lines joining the top of each pole to the foot of the other is a' metros. Find 'a'

In $△ABC$ and $△QPC$

$\angle BCA=\angle PCQ$ (common)

$\angle CPQ=\angle CBA$ (right angle)

So, $△ABC\sim △QPC$ (AA criteria)

Using BPT,

$\frac{PQ}{AB}=\frac{CP}{BC}⟶i$

In $△CBD$ and $△PBQ$

$\angle CBD$ is common,

and $\angle BPQ=\angle BCD$ (right angle)

So, $△CBD=△PBQ$ (AAA criteria)

Using BPT, $\frac{PQ}{CD}=\frac{BP}{BC}⟶ii$

Adding $i$ and $ii$ we get,

$\frac{PQ}{AB}+\frac{PQ}{CD}=\frac{CP+BP}{BC}=1$

$\frac{a}{20}+\frac{a}{30}=1$

$a=16m$