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
(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:
$\angle B A C=\angle E D F=90^{\circ}$
$\angle A C B=\angle D F E \quad$ (Angular elevation of the Sun at the same time)
Therefore, by AA similarity theorem, we get:
$\triangle A B C \sim \triangle D E F$
$\Rightarrow \frac{A B}{A C}=\frac{D E}{D F}$
$\Rightarrow \frac{6}{3.6}=\frac{D E}{18}$
$\Rightarrow D E=\frac{6 \times 18}{36}=30 \mathrm{~m}$
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