Question

A vertical pole 6 m long casts a shadow of length 3.6 m on the ground. What is the height of a tower which casts a shadow of length 18 m at the same time?
(a) 10.8 m
(b) 28.8 m
(c) 32.4 m
(d) 30 m

Answer 1

(d) 30 m

Let AB and AC be the vertical pole and its shadow, respectively.
According to the question:
AB = 6 m
AC = 3.6 m
Again, let DE and DF be the tower and its shadow.
According to the question:
DF = 18 m
​DE = ?
Now, in right-angled triangles ABC and DEF, we have:
$$\begin{gathered} \angle BAC=\angle EDF=90° \\ \angle ACB=\angle DFE(\mathrm{Angular}\mathrm{elevation}\mathrm{of}\mathrm{the}\mathrm{Sun}\mathrm{at}\mathrm{the}\mathrm{same}\mathrm{time}) \\ \mathrm{Therefore},\mathrm{by}\mathrm{AA}\mathrm{similarity}\mathrm{theorem},\mathrm{we}\mathrm{get}: \\ △ABC~△DEF \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{AB}{AC}=\frac{DE}{DF} \\ \Rightarrow \frac{6}{3.6}=\frac{DE}{18} \\ \Rightarrow DE=\frac{6\times 18}{3.6}=30\mathrm{m} \end{gathered}$$