The horizontal distance between two poles is $ 15 \mathrm{~m} $. The angle of depression of the top of the first pole as seen from the top of the second pole is $ 30^{\circ} $. If the height of the second pole is $ 24 \mathrm{~m} $, find the height of the first pole. $ (\sqrt{3}=1.732) \quad $
Given:
The horizontal distance between two poles is \( 15 \mathrm{~m} \).
The angle of depression of the top of the first pole as seen from the top of the second pole is \( 30^{\circ} \).
The height of the second pole is \( 24 \mathrm{~m} \)
To do:
We have to find the height of the first pole.
Solution:
Let $AB$ and $CD$ be the two poles, the height of the second pole $CD=24\ m$.
Let the height of first pole $AB=h\ m$
Distance between the two poles, $AC = 15\ m$
From the figure,
$BE=AC= 15\ m$ and $AB=EC=h$ and $DE = 24-h\ m$
In $\vartriangle DBE$,
$tan30^{o}=\frac{DE}{BE}$
$\Rightarrow \frac{1}{\sqrt{3}} =\frac{24-h}{15}$
$\Rightarrow 24-h=\frac{15}{\sqrt{3}}$
$\Rightarrow h=24-\ \frac{15}{\sqrt{3}}$
$\Rightarrow h=24-5\sqrt{3}$
$\Rightarrow h=24-5\times 1.732$
$h=15.34\ m$
Therefore, the height of the first pole is $15.34\ m$.
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