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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.
Solution
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.
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