From the top of a building $ AB, 60 \mathrm{~m} $ high, the angles of depression of the top and bottom of a vertical lamp post $ CD $ are observed to be $ 30^{\circ} $ and $ 60^{\circ} $ respectively. Find the difference between the heights of the building and the lamp post.

Given:

From the top of a building \( A B, 60 \mathrm{~m} \) high, the angles of depression of the top and bottom of a vertical lamp post \( C D \) are observed to be \( 30^{\circ} \) and \( 60^{\circ} \) respectively.

To do:

We have to find the difference between the heights of the building and the lamp post.

Solution:  

From the figure,

$\mathrm{AB}=60 \mathrm{~m}, \angle \mathrm{BDE}=30^{\circ}, \angle \mathrm{BCA}=60^{\circ}$

Let the horizontal distance between \( A B \) and \( C D \) be $\mathrm{AC}=x \mathrm{~m}$ and the height of the lamp post be $\mathrm{CD}=h \mathrm{~m}$.

This implies,

$\mathrm{AE}=\mathrm{CD}=h \mathrm{~m}$

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

$\mathrm{BE}=60-h \mathrm{~m}$

We know that,

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

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

$\Rightarrow \tan 30^{\circ}=\frac{60-h}{x}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{60-h}{x}$

$\Rightarrow x=(60-h)\sqrt3 \mathrm{~m}$............(i)

Similarly,

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

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

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

$\Rightarrow \sqrt3=\frac{60}{(60-h)\sqrt3}$                  [From (i)]

$\Rightarrow [(60-h)\sqrt3]\sqrt3=60 \mathrm{~m}$

$\Rightarrow (60-h)3=60 \mathrm{~m}$

$\Rightarrow 60-h=20 \mathrm{~m}$

$\Rightarrow h=60-20 \mathrm{~m}$

$\Rightarrow h=40 \mathrm{~m}$

$\Rightarrow x=(60-40)(1.73)=20(1.73)=34.64 \mathrm{~m}$

$\Rightarrow \mathrm{AB}-\mathrm{CD}=60-40=20 \mathrm{~m}$

Therefore, the difference between the heights of the building and the lamp post is $20 \mathrm{~m}$.    

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Updated on: 10-Oct-2022

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