Question

An aeroplane at an altitude of 200 metres observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. Find the width of the river. (Use = `sqrt(3)` = 1.732)

Answer 1

Let the aeroplane's position be A; B and C be two points on the two banks of a river such that the angles of depression at B and C are 45° and 60° respectively.

Let BD = x m, CD = y m

Given, AD = 200 m

In ΔADB, ∠D = 90°

tan 45° = `(AD)/(BD)`

⇒ 1 = `200/x`

⇒ x = 200 m  ...(i)

In ΔADC, ∠D = 90°

tan 60° = `(AD)/(CD)`

⇒ `sqrt(3) = 200/y`

⇒ y = `200/sqrt(3)`

⇒ y = `(200sqrt(3))/3`  ...(ii)

On adding equations (i) and (ii), we get

x + y = `200 + (200sqrt(3))/3`

= `(600 + 200sqrt(3))/3`

= `(200(3 + sqrt(3)))/3`

= `(200(3 + 1.732))/3`

= `(200 xx 4.732)/3`

= `946.4/3`

= 315.4 m

Hence, the width of the river is 315.4 m.