A man sitting at a height of $ 20 \mathrm{~m} $ on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with the foot of tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are $ 60^{\circ} $ and $ 30^{\circ} $ respectively. Find the width of the river.
Given:
A man sitting at a height of \( 20 \mathrm{~m} \) on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with the foot of tree.
The angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are \( 60^{\circ} \) and \( 30^{\circ} \) respectively.
To do:
We have to find the width of the river.
Solution:
Let $AB$ be the height of the tree and $C, D$ are the points of depression of the banks on the opposite side of the river.
From the figure,
$\mathrm{AB}=20 \mathrm{~m}, \angle \mathrm{ACB}=60^{\circ}, \angle \mathrm{ADB}=30^{\circ}$
Let the distance between the tree and point $C$ be $\mathrm{BC}=x \mathrm{~m}$ and the distance between the tree and point $D$ be $\mathrm{BD}=y \mathrm{~m}$.
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { AB }}{BC}$
$\Rightarrow \tan 60^{\circ}=\frac{20}{x}$
$\Rightarrow \sqrt3=\frac{20}{x}$
$\Rightarrow x=\frac{20}{\sqrt3} \mathrm{~m}$.........(i)
Similarly,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { AB }}{BD}$
$\Rightarrow \tan 30^{\circ}=\frac{20}{y}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{20}{y}$
$\Rightarrow y=20\sqrt3 \mathrm{~m}$..........(ii)
$\Rightarrow x+y=\frac{20}{\sqrt3}+20\sqrt3 \mathrm{~m}$
$\Rightarrow x+y=\frac{20+20(3)}{\sqrt3} \mathrm{~m}$
$\Rightarrow x+y=\frac{80}{\sqrt3} \mathrm{~m}$
Therefore, the width of the river is $\frac{80}{\sqrt3} \mathrm{~m}$.
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