Question

A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°.

Calculate :

1. the width of the river;
2. the height of the tree.

Answer 1

Let AB be the tree and AC be the width of the river.

Let D be a point such that CD = 50 m.

Given that ∠BCA = 60° and ∠BDA = 30° 

In ΔBAD, 

`(BA)/(AD) = tan 30^circ`

`=> BA = (AD)/sqrt(3)`  ...(i)

In ΔBAC,

`(BA)/(AC) = tan 60^circ`

`=> BA = ACsqrt(3)`  ...(ii)

From (i) and (ii), we get 

`(AD)/(sqrt(3)) = ACsqrt(3)`

`=>` (50 + AC) = 3AC

∴ AC = 25 m

Thus, width of the river is 25 m.

From (ii),

BA = 25 × 1.732 = 43.3 m 

Hence, height of the tree is 43.3 m.

Answer 2

Let H be the height of the tree

Let height of the tree be H meter.

In right-angled ΔACD,

`tan 60^circ = H/(CD)` 

`sqrt(3) = H/(CD)` 

∴ `CD = H/sqrt(3)`   ...(i)

In right-angled ΔABD,

`tan 30^circ = H/(BD)`

∴ `1/sqrt(3) = H/(BD)`

`BD = sqrt(3)H`     ...(ii)

BD – CD = 50

`(sqrt(3)H)/1 - H/sqrt(3) = 50`    ...(Using (i) and (ii))

∴ `(3H - H )/sqrt(3) = 50`

∴ `2H = 50sqrt(3)`

or `H = (50sqrt(3))/2 = 25sqrt(3)`

H = 43.3 m

i. The width of the river `CD = (25sqrt(3))/sqrt(3) = 25  m`

ii. The height of the tree H = 43.3 m.