The angle of elevation of the top of a building from the foot of a tower is $30^o$ and the angle of elevation of the top of the tower from the foot of the building is $60^o$. If the tower is $50\ m$ high, find the height of the building.
Given:
The angle of elevation of the top of the building from the foot of the tower is \( 30^{\circ} \) and the angle of the top of the tower from the foot of the building is \( 60^{\circ} \).
The tower is \( 50 \mathrm{~m} \) high.
To do:
We have to find the height of the building.
Solution:
Let $AB$ be the height of the tower and $CD$ be the height of the building.
From the figure,
$\mathrm{AB}=50 \mathrm{~m}, \angle \mathrm{BCA}=60^{\circ}, \angle \mathrm{DAC}=30^{\circ}$
Let the height of the building be $\mathrm{CD}=h \mathrm{~m}$ and the distance between the building and the tower be $\mathrm{CA}=x \mathrm{~m}$.
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { AB }}{CA}$
$\Rightarrow \tan 60^{\circ}=\frac{50}{x}$
$\Rightarrow \sqrt3=\frac{50}{x}$
$\Rightarrow x=\frac{50}{\sqrt3} \mathrm{~m}$.........(i)
Similarly,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { DC }}{AC}$
$\Rightarrow \tan 30^{\circ}=\frac{h}{x}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{h}{\frac{50}{\sqrt3}}$ [From (i)]
$\Rightarrow \frac{1}{\sqrt3}\times\frac{50}{\sqrt3}=h \mathrm{~m}$
$\Rightarrow h=\frac{50}{3} \mathrm{~m}$
Therefore, the height of the building is $\frac{50}{3} \mathrm{~m}$.
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