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A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Given:
A vertical pole $6\ m$ long casts a shadow $4\ m$ long.
A tower casts a shadow $28\ m$ long.
To do:
We have to find the height of the tower.
Solution:
Length of the pole $= 6\ m$.
Length of the pole’s shadow $= 4\ m$.
Length of the tower’s shadow $= 28\ m$.
Let the height of the tower be $h\ m$.
In $\triangle ABC$ and $\triangle PQR$
$\angle ABC = \angle PQR = 90^o$
$\angle ACB = \angle PRQ$ (Angular elevation of Sun is same)
Therefore,
$\triangle ABC \sim\ \triangle PQR$ (By AA similarity)
This implies,
$\frac{AB}{BC} = \frac{PQ}{QR}$ (Corresponding sides are proportional)
$\frac{6}{4} = \frac{h}{28}$
$h = \frac{28\times6}{4}$
$h = 7\times6$
$h = 42\ m$
The height of the tower is $42\ m$.
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