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. Find the horizontal distance between \( A B \) and \( C D \).
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 horizontal distance between \( A B \) and \( C D \).
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)]
$\Rightarrow [(60-h)\sqrt3]\sqrt3=60 \mathrm{~m}$
$\Rightarrow (60-h)3=60 \mathrm{~m}$
$\Rightarrow 60-h=20 \mathrm{~m}$
$\Rightarrow h=60-20 \mathrm{~m}$
$\Rightarrow h=40 \mathrm{~m}$
$\Rightarrow x=(60-40)(1.73)=20(1.73)=34.64 \mathrm{~m}$
Therefore, the horizontal distance between $AB$ and $CD$ is $34.64 \mathrm{~m}$.
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