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Question
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)`
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
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