The angle of elevation of the top of a tower from a point $ A $ on the ground is $ 30^{\circ} $. On moving a distance of 20 metres towards the foot of the tower to a point $ B $ the angle of elevation increases to $ 60^{\circ} $. Find the height of the tower and the distance of the tower from the point $ A $.

Given:

The angle of elevation of the top of a tower from a point \( A \) on the ground is \( 30^{\circ} \). On moving a distance of 20 metres towards the foot of the tower to a point \( B \) the angle of elevation increases to \( 60^{\circ} \).

To do:

We have to find the height of the tower and the distance of the tower from the point \( A \).

Solution:  

Let $CD$ be the tower and $AB$ be the distance moved towards the foot of the tower starting from $A$.

From the figure,

$\mathrm{AB}=20 \mathrm{~m}, \angle \mathrm{CAD}=30^{\circ}, \angle \mathrm{CBD}=60^{\circ}$

Let the height of the tower be $\mathrm{CD}=h \mathrm{~m}$ and the distance between the point $A$ and the foot of the tower be $\mathrm{AD}=x \mathrm{~m}$.

This implies,

$\mathrm{BD}=x-20 \mathrm{~m}$

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { CD }}{AD}$

$\Rightarrow \tan 30^{\circ}=\frac{h}{x}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{h}{x}$

$\Rightarrow x=h(\sqrt3) \mathrm{~m}$

$\Rightarrow x=\sqrt3 h \mathrm{~m}$...........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { CD }}{DB}$

$\Rightarrow \tan 60^{\circ}=\frac{h}{x-20}$

$\Rightarrow \sqrt3=\frac{h}{x-20}$

$\Rightarrow (x-20)\sqrt3=h \mathrm{~m}$

$\Rightarrow (\sqrt3 h-20)\sqrt3=h \mathrm{~m}$           [From (i)]

$\Rightarrow 3h-h=20(1.732) \mathrm{~m}$

$\Rightarrow 2h=34.64 \mathrm{~m}$

$\Rightarrow h=\frac{34.64}{2}=17.32 \mathrm{~m}$

$\Rightarrow x=1.732(17.32)=30 \mathrm{~m}$

Therefore, the height of the tower is $17.3 \mathrm{~m}$ and the distance of the first position from the tower is $30 \mathrm{~m}$. 

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Updated on: 10-Oct-2022

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