The angle of elevation of a tower from a point on the same level as the foot of the tower is $ 30^{\circ} $. On advancing 150 metres towards the foot of the tower, the angle of elevation of the tower becomes $ 60^{\circ} $. Show that the height of the tower is $ 129.9 $ metres (Use $ \sqrt{3}=1.732 $ ).

Given:

The angle of elevation of a tower from a point on the same level as the foot of the tower is \( 30^{\circ} \). On advancing 150 metres towards the foot of the tower, the angle of elevation of the tower becomes \( 60^{\circ} \).

To do:

We have to show that the height of the tower is \( 129.9 \) metres.

Solution:  

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

From the figure,

$\mathrm{CD}=150 \mathrm{~m}, \angle \mathrm{ACB}=30^{\circ}, \angle \mathrm{ADB}=60^{\circ}$

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,

$\mathrm{DB}=x-150 \mathrm{~m}$

We know that,

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

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

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

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

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

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

Similarly,

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

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

$\Rightarrow \tan 60^{\circ}=\frac{h}{x-150}$

$\Rightarrow \sqrt3=\frac{h}{x-150}$

$\Rightarrow (x-150)\sqrt3=h \mathrm{~m}$

$\Rightarrow (\sqrt3 h-150)\sqrt3=h \mathrm{~m}$           [From (i)]

$\Rightarrow 3h-150\sqrt3=h \mathrm{~m}$

$\Rightarrow 3h-h=150\sqrt3 \mathrm{~m}$

$\Rightarrow h=\frac{150\times1.732}{2}=129.9 \mathrm{~m}$

Therefore, the height of the tower is $129.9 \mathrm{~m}$.   

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Updated on: 10-Oct-2022

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