From the top of a $ 50 \mathrm{~m} $ high tower, the angles of depression of the top and bottom of a pole are observed to be $ 45^{\circ} $ and $ 60^{\circ} $ respectively. Find the height of the pole..

Given:

From the top of a \( 50 \mathrm{~m} \) high tower, the angles of depression of the top and bottom of a pole are observed to be \( 45^{\circ} \) and \( 60^{\circ} \) respectively. 

To do:

We have to find the height of the pole.

Solution:  

Let $AB$ be the tower and $CD$ be the pole.

Let $B$ be the point of observation.

From the figure,

$\mathrm{AB}=50 \mathrm{~m}, \angle \mathrm{BDE}=45^{\circ}, \angle \mathrm{BCA}=60^{\circ}$

Let the height of the pole be $\mathrm{CD}=h \mathrm{~m}$ and the distance between the pole and the tower be $\mathrm{AC}=x \mathrm{~m}$.

This implies,

$\mathrm{AE}=50-h \mathrm{~m}$

$\mathrm{AC}=\mathrm{DE}=x \mathrm{~m}$

We know that,

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

$=\frac{\text { BE }}{DE}$

$\Rightarrow \tan 45^{\circ}=\frac{50-h}{x}$

$\Rightarrow 1(x)=50-h$

$\Rightarrow x=50-h \mathrm{~m}$............(i)

Similarly,

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

$=\frac{\text { BA }}{CA}$

$\Rightarrow \tan 60^{\circ}=\frac{50}{x}$

$\Rightarrow \sqrt3=\frac{50}{50-h}$                  [From (i)]

$\Rightarrow (50-h)\sqrt3=50 \mathrm{~m}$

$\Rightarrow h\sqrt3=50(\sqrt3-1) \mathrm{~m}$

$\Rightarrow h(1.732)=50(1.732-1) \mathrm{~m}$

$\Rightarrow h=\frac{50(0.732)}{1.732} \mathrm{~m}$

$\Rightarrow h=\frac{36.6}{1.732} \mathrm{~m}$

$\Rightarrow h=21.13 \mathrm{~m}$

Therefore, the height of the pole is $21.13 \mathrm{~m}$.   

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Updated on: 10-Oct-2022

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