A person in a helicopter flying at a height of $500$ m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are $30^{o}$ and $45^{o}$ . Find the width of the river. $(\sqrt{3} = 1.732)$

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Solution

Let $A B$ be the width of the river.
$A$ and $B$ are two points on the opposite water level.
Let $C$ be the helicopter.
Let $A D = x$ , $B D = y$
$Δ A C D$ and $Δ B C D$ ,
$\Rightarrow tan 30^{o} = \frac{C D}{A D}$ and $tan 45^{o} = \frac{C D}{D B}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{500}{x}$ and $1 = \frac{500}{y}$

$\Rightarrow x = 500 \sqrt{3}$ and $y = 500$
The width of river $= x + y$
$= 500 \sqrt{3} + 500$
$= 500 (\sqrt{3} + 1)$
$= 500 (1.732 + 1)$
$= 500 (2.732)$
$= 1366.000$
$= 1366$ m

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A person in a helicopter flying at a height of 500 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30o and 45o. Find the width of the river. (√3=1.732)

A person in a helicopter flying at a height of $500$ m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are $30^{o}$ and $45^{o}$ . Find the width of the river. $(\sqrt{3} = 1.732)$