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A person standing on the bank of river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60° - Geometry

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Sum

A person standing on the bank of river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and width of the river. `(sqrt 3=1.73)`

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Solution 1

Let AB = height of the tower = h metres
In the right angled ΔABC and right angled ΔABD,

`tan60^@=h/(BC)=sqrt(3) rArr BC=h/sqrt(3)`

`tan30^@=h/(BD)=1/sqrt(3) rArr BD=hsqrt(3)`

`Now, BD – BC = 40`

`hsqrt(3)-h/sqrt(3)=40`

`(3h-h)/sqrt3=40`

`2h=40sqrt3`

`h=20sqrt3 m`

In right angled ΔABC,

`tan30°=1/sqrt3`

`(BC)/h=1/sqrt3`

`(BC)/(20sqrt3)=1/sqrt3`

`BC=b=20m`

Width of the river = BC = 20 m
Thus, the height of the tree is `20 sqrt3` metres and width of the river is 20 metres.

Solution 2

Let BC be the height of the tree.

AB be the breadth of the river.

A be the initial position of the person

D be the final position of the person

CAB = 60° & CDB = 30° & DA = 40m

Let AB = x & BC = h

In Δ DBC

⇒ `tan 30 = "BC"/"DB" = "BC"/"DA+AB" ="h"/(40+"x")`

⇒ `1/sqrt3 = "h"/(40+"x")`

⇒ `"h" = 40+"x"/sqrt3` ................(1)

In Δ ABC

⇒ `tan60  = "BC"/"AB"="h"/"x"`

⇒ `sqrt3 = "h"/"x"`

⇒ `"h"=sqrt3 "x"`   ..................(2)

Using (1) & (2)

`"h"=(40+"x")/sqrt3`

⇒ `sqrt(3"x")=(40+"x")/sqrt3`

⇒ 3x =40 + x

⇒ 2x = 40

⇒ x = 20

 AB = 20m

 BC = `20sqrt3` m =20 ×1.73 =34.6m

Height of the tree is 34.6m and the width of the river is 20m.

Solution 3

Consider the above diagram where AB is the height of a tree with point A as the top of the tree

Point B base of the tree

Initially, the person is at point C, therefore, BC is the width of the river and the person observes the angle of elevation to be 60° i.e. ACB = 60°

The person moves 40 m away from the bank of the rives thus the new position of the person is point D and CD = 40m

From D the person observes the angle of elevation to be 30° i.e. ADB = 30°

Consider ∆ABC

tan60° = `"AB"/"BC"`

AB = `"BC"sqrt3` …(i)

Consider ∆ABD

tan30° = `"AB"/"BD"`

`1/sqrt(3) = "AB"/("BC + CD")`

Using CD = 40 m

(BC + 40) = `"AB"sqrt3`

 AB = `("BC" + 40)/sqrt3` …(ii)

From (i) and (ii)

`"BC"sqrt3` = `("BC" + 40)/sqrt3`

`"BC"sqrt(3) xx sqrt(3) = "BC" + 40`

 3BC = BC + 40

 2BC = 40

 BC = 20 m

Therefore width of the river = BC = 20 m

Substituting BC in equation (i)

AB = `20sqrt3`

 AB = 20 × 1.73

 AB = 34.6 m

Therefore height of tree = AB = 34.6 m

Hence height of tree is 34.6 meters and width of river is 20 meters

Concept: Heights and Distances
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