Question

At a point on level ground, the angle of elevation of a vertical tower is, found to be α such that tan α = `1/3`. After walking 100 m towards the tower, the angle of elevation β becomes such that tan β = `3/4`. Find the height of the tower. 

Answer 1

Let the height of the tower be h m.

Given, tan α = `1/3` and tan β = `3/4`

Now, In ΔABD, ∠D = α, ∠B = 90°

 tan α = `("AB")/("DB")`

⇒ `1/3 =  "h"/(100 + x)`

⇒ 100 + x = 3h

⇒ x = 3h – 100  ...(i)

In ΔABC, ∠C = β, ∠B = 90°

tan β = `("AB")/("BC")`

⇒ `3/4 = "h"/x`

⇒ x = `(4"h")/3`  ...(ii)

From equations (i) and (ii), we have

3h – 100  = `(4"h")/3`

⇒ `3"h" - (4"h")/3` = 100

⇒ `(5"h")/3` = 100

⇒ h = `300/5`

⇒ h = 60 m

Hence, the height of the tower is 60 m.