Question

A vertical pole of length $6m$ casts a shadow $4m$ long on the ground and at the same time a tower casts a shadow $28m$ long. Find the height of the tower.

Answer 1

Let $AB$ be the pole and $BC$ be its shadow. At the same time let $PQ$ be the tower and $QR$ be its shadow.

i.e., $AB=6m$, $BC=4m$ and $QR=28m$

Practically when sunlight falls on pole AB, then the shadow BC is created. The same is with the case of Tower PQ. But in this case, the angle of elevation of shadow with the sun will be the same in both the cases i.e.,
$\angle BAC=\angle QPR..........\left(1\right)$

In $△ABC$ and $△PQR$
$\angle B=\angle Q={90}^o$ [ The objects and shadows are perpendicular to each other]
$\therefore$ By AA Criterion of Similarity $△ABC$ $\sim$ $△PQR$

The ratio of any two corresponding sides in two equiangular triangles is always the same

$\therefore$ $\frac{AB}{PQ}=\frac{BC}{QR}$

$\therefore$ $\frac{6}{PQ}=\frac{4}{28}$

$\therefore$ $PQ=42m$

So, the height of the tower is $42$ m.