Question

A vertical stick of length 7.5 m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower.

Answer 1

Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m

Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m.

In $△ABC\mathrm{and}△PQR$, we have:
$$\begin{gathered} \angle ABC=\angle PQR=90° \\ \angle ACB=\angle PRQ(\mathrm{Angular}\mathrm{elevation}\mathrm{of}\mathrm{the}\mathrm{Sun}\mathrm{at}\mathrm{the}\mathrm{same}\mathrm{time}) \end{gathered}$$
Therefore, by AA similarity theorem, we get:
$△ABC~△PQR$
$\Rightarrow \frac{AB}{BC}=\frac{PQ}{QR}$

$$\begin{gathered} \Rightarrow \frac{7.5}{5}=\frac{x}{24} \\ \Rightarrow x=\frac{7.5}{5}\times 24=36\mathrm{m} \end{gathered}$$
Therefore, PQ = 36 m

Hence, the height of the tower is 36 m.