A person in a helicopter flying at a height of $700$ 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. $($ $\sqrt{3}$ $= 1.732$ $)$

Open in App

Solution

Let $C$ be the position of a person in a helicopter. $A, B$ be two objects lying opposite to each other on either bank of a river.
Let $A D = x, B D = y$
In $Δ A C D$ ,
$\Rightarrow tan 30^{o} = \frac{C D}{A D}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{700}{x}$
$\Rightarrow x = 700 \sqrt{3}$
In $Δ B C D$ ,
$tan 45^{\circ} = \frac{C D}{B D}$
$⟹ B D = C D$
$⟹ B D = y = 700$
$∴$ the width of the river $= x + y$
$⟹ 700 \sqrt{3} + 700 m$

Suggest Corrections

Similar questions

Q. A helicopter is flying at an altitude of

250 m

between two banks of a river. From the helicopter it is observed that the angles of a river. From the helicopter its is observed that the angle of depression of two boats on the opposite banks are

45^{o}

and

60^{o}

respectively. Find the width of the river. Given your answer correct to nearest metre.

An aeroplane is flying at a height of 300 m above the gound. Flying at this height the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are $45^{\circ}$ and $60^{\circ}$ respectively. Find the width of the river. [Use~ $\sqrt{3} = 1.732.$ ]

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^{\circ} a n d 45^{\circ}$ respectively. If the bridge is at a height of 2.5 m from the banks, find the width of the river. [Take $\sqrt{3} = 1.732.$ ]

Join BYJU'S Learning Program

A person in a helicopter flying at a height of 700 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 )

Let C be the position of a person in a helicopter. A,B be two objects lying opposite to each other on either bank of a river.Let AD=x,BD=yIn ΔACD,⇒tan30o=CDAD⇒1√3=700x⇒x=700√3In ΔBCD,tan45∘=CDBD⟹BD=CD⟹BD=y=700∴ the width of the river =x+y⟹700√3+700 m

A person in a helicopter flying at a height of $700$ 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. $($ $\sqrt{3}$ $= 1.732$ $)$