The angle of elevation of the top of a chimney from the top of a tower is $ 60^{\circ} $ and the angle of depression of the foot of the chimney from the top of the tower is $ 30^{\circ} $. If the height of the tower is $ 40 \mathrm{~m} $, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be $ 100 \mathrm{~m} $. State if the height of the above mentioned chimney meets the polution norms. What value is discussed in this question?
Given:
The angle of elevation of the top of a chimney from the top of a tower is \( 60^{\circ} \) and the angle of depression of the foot of the chimney from the top of the tower is \( 30^{\circ} \).
The height of the tower is \( 40 \mathrm{~m} \).
According to pollution control norms, the minimum height of a smoke emitting chimney should be \( 100 \mathrm{~m} \).
To do:
We have to find the height of the chimney and state whether the height of the above mentioned chimney meets the pollution norms.
Solution:
Let $AB$ be the tower and $CD$ be the chimney.
From the figure,
$\angle DBE=60^{\circ}, \angle EBC=\angle BCA=30^{\circ}, \mathrm{AB}=40\ m$
Let the height of the chimney be $\mathrm{CD}=h \mathrm{~m}$ and the distance between the chimney and the tower be $\mathrm{AC}=x \mathrm{~m}$
This implies,
$\mathrm{EC}=\mathrm{AB}=40\ m$
$\mathrm{DE}=h-40\ m$
$\mathrm{BE}=\mathrm{AC}=x \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { AB }}{AC}$
$\Rightarrow \tan 30^{\circ}=\frac{40}{x}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{40}{x}$
$\Rightarrow x=40(\sqrt3)$
$\Rightarrow x=40\sqrt3$.............(i)
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { DE }}{BE}$
$\Rightarrow \tan 60^{\circ}=\frac{h-40}{x}$
$\Rightarrow \sqrt3=\frac{h-40}{x}$
$\Rightarrow x\sqrt3=h-40$
$\Rightarrow 40\sqrt3(\sqrt3)=h-40$
$\Rightarrow 40(3)+40=h$
$\Rightarrow h=160\ m$
Therefore, the height of the chimney is $160\ m$ and it meets the pollution norms.
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