The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is $ 32^{\circ} $. When the observer moves towards the tower a distance of $ 100 \mathrm{~m} $, he finds the angle of elevation of the top to be $ 63^{\circ} $. Find the height of the tower and the distance of the first position from the tower. [Take $ \tan 32^{\circ}=0.6248 $ and tan $ \left.63^{\circ}=1.9626\right] $

Given:

The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is \( 32^{\circ} \).

When the observer moves towards the tower a distance of \( 100 \mathrm{~m} \), he finds the angle of elevation of the top to be \( 63^{\circ} \).

To do:

We have to find the height of the tower and the distance of the first position from the tower.

Solution:  

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

From the figure,

$\mathrm{CD}=100 \mathrm{~m}, \angle \mathrm{ACB}=32^{\circ}, \angle \mathrm{ADB}=63^{\circ}$

Let the height of the tower be $\mathrm{AB}=h \mathrm{~m}$ and the distance between the point $C$ and the foot of the tower be $\mathrm{BC}=x \mathrm{~m}$.

This implies,

$\mathrm{DB}=x-100 \mathrm{~m}$

We know that,

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

$=\frac{\text { AB }}{BC}$

$\Rightarrow \tan 32^{\circ}=\frac{h}{x}$

$\Rightarrow 0.6248=\frac{h}{x}$

$\Rightarrow x(0.6248)=h \mathrm{~m}$

$\Rightarrow h=0.6248x \mathrm{~m}$...........(i)

Similarly,

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

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

$\Rightarrow \tan 63^{\circ}=\frac{h}{x-100}$

$\Rightarrow 1.9626=\frac{h}{x-100}$

$\Rightarrow (x-100)1.9626=h \mathrm{~m}$

$\Rightarrow 1.9626x-196.26=0.6248x \mathrm{~m}$           [From (i)]

$\Rightarrow (1.9626-0.6248)x=196.26 \mathrm{~m}$

$\Rightarrow 1.3378x=196.26 \mathrm{~m}$

$\Rightarrow x=\frac{196.26}{1.3378}=146.70 \mathrm{~m}$

$\Rightarrow h=0.6248(146.70)=91.65 \mathrm{~m}$

Therefore, the height of the tower is $91.65 \mathrm{~m}$ and the distance of the first position from the tower is $146.70 \mathrm{~m}$.

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Updated on: 10-Oct-2022

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