Question

To find the width of the river, a man observes the top of a tree, on the opposite bank making an angle of elevation of 60°. When he moves 24 metres backward from bank and observes the same top of the tree, his line of vision makes an angle of elevation of 30°. Find the height of the tree and width of the river. (`sqrt3 = 1.73`)

Answer 1

Let AB represent the tree on the opposite bank of the river.

CB represents the width of the river. 

C is the initial position of the observer.

D is the final position of the observer.

DC = 24 m.

Let AB be x m and BC be y m.

∠ BCA = 60° and ∠ BDA = 30°.

In △ABC,

tan ∠ BCA = tan 60° = `("AB")/("BC")`

∴ `sqrt3 = x/y`

∴ x = `sqrt3`y

In ΔABC,

tan ∠BCA = tan 60° = `("AB")/("BC")`

∴ `sqrt3 = x/y` 

∴ x = `sqrt3`y           ...(1)

ln △ABD,

tan ∠ADB = tan 30° = `("AB")/("BD")`

∴ `1/sqrt3 = x/(24 + y)`

∴ x = `(24 + y)/sqrt3`          ...(2)

From (1) and (2),

`sqrt3y = (24 + y)/sqrt3`

∴ `sqrt3 y xx sqrt3 = 24 + y`

∴ 3y = 24 + y

∴ 3y − y + 24

∴ 2y = 24

∴ y = 12       ...(Width of the river)   ...(3)

The height of the tree = x m

`sqrt3 xx 12` m        ...[From (1) and (3)]

= 1.73 × 12 m

= 20.76 m

The height of the tree is 20.76 m and the width of the river is 12 m.