Question

A 15 high tower casts a shadow 24 long at a certain time and at the same time, a telephone pole casts a shadow 16 long. Find the height of the telephone pole.

Answer 1

Let BC = 15 m be the tower and its shadow AB is 24 m. at that time $\angle CAB=\theta$. Again, let EF = h be a telephone pole and its shadow DE = 16m. At the same time $\angle EDF=\theta$.
Here, $\Delta ABC$ and $\Delta DEF$ both are right angled triangles.



In $\Delta ABC$ and $\Delta DEF$, $\angle CAB=\angle EDF=\theta$
$\angle B=\angle E$ $\left[each{90}^{\circ }\right]$
$\therefore$ $ABC\sim DEF$ [ by AAA similary criterion]
Then, $\frac{AB}{DE}=\frac{BC}{EF}$
$\Rightarrow$ $\frac{24}{16}=\frac{15}{h}$
$\therefore$ $h=\frac{15\times 16}{24}=10$
Hence, the height of the telephone pole is 10m.