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Question
The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.
Solution
Let distance between the two towers = AB = x m
And height of the other tower = PA = h m
Given, height of the tower = QB = 30 m, ∠QAB = 60° and ∠PBA = 30°
Now, in ∆QAB,
tan 60° = `"QB"/"AB" = 30/x`
⇒ `sqrt(3) = 30/x`
⇒ `x = 30/sqrt(3)`
⇒ `x = 30/sqrt(3) * sqrt(3)/sqrt(3)`
= `(30sqrt(3))/3`
= `10sqrt(3)`
And in ∆PBA,
tan 30° = `"PA"/"AB" = "h"/x`
⇒ `1/sqrt(3) = "h"/(10sqrt(3))` ...`[∵ x = 10sqrt(3) "m"]`
⇒ h = 10
Hence, the required distance and height are `10sqrt(3) "m"` and 10 m, respectively.
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