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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)
Solution
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.
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