# 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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