From the top of a building $ 15 \mathrm{~m} $ high the angle of elevation of the top of a tower is found to be $ 30^{\circ} $. From the bottom of the same building, the angle of elevation of the top of the tower is found to be $ 60^{\circ} $. Find the height of the tower and the distance between the tower and building.
Given:
From the top of a building \( 15 \mathrm{~m} \) high the angle of elevation of the top of a tower is found to be \( 30^{\circ} \).
From the bottom of the same building, the angle of elevation of the top of the tower is found to be \( 60^{\circ} \).
To do:
We have to find the height of the tower and the distance between the tower and building.
Solution:
Let $CD$ be the building and $AB$ be the tower.
From the figure,
$\mathrm{AB}=15 \mathrm{~m}, \angle \mathrm{CAE}=30^{\circ}, \angle \mathrm{CBD}=60^{\circ}$
Let the height of the tower be $\mathrm{CD}=h \mathrm{~m}$ and the distance between the tower and the building be $\mathrm{BD}=x \mathrm{~m}$.
This implies,
$\mathrm{AE}=\mathrm{BD}=x \mathrm{~m}$
$\mathrm{ED}=\mathrm{AB}=15 \mathrm{~m}$ and $\mathrm{CE}=h-15 \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { CE }}{AE}$
$\Rightarrow \tan 30^{\circ}=\frac{h-15}{x}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{h-15}{x}$
$\Rightarrow x=(h-15)(\sqrt3) \mathrm{~m}$
$\Rightarrow x=\sqrt3 (h-15) \mathrm{~m}$...........(i)
Similarly,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { CD }}{DB}$
$\Rightarrow \tan 60^{\circ}=\frac{h}{x}$
$\Rightarrow \sqrt3=\frac{h}{x}$
$\Rightarrow (x)\sqrt3=h \mathrm{~m}$
$\Rightarrow [\sqrt3 (h-15)]\sqrt3=h \mathrm{~m}$ [From (i)]
$\Rightarrow 3(h-15)=h \mathrm{~m}$
$\Rightarrow 3h-h=45 \mathrm{~m}$
$\Rightarrow h=\frac{45}{2}=22.5 \mathrm{~m}$
$\Rightarrow x=1.732(22.5-15)=1.732(7.5)=12.975 \mathrm{~m}$
Therefore, the height of the tower is $22.5 \mathrm{~m}$ and the distance between the tower and building is $12.975 \mathrm{~m}$.
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