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Question
The angle of depression form the top of a tower of a point A on the ground is 30° . On moving a distance of 20 meters from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower to from the point B is 60° . Find the height of the tower and its distance from the point A.
Solution
Let PQ be the tower.
We have,
AB = 20m, ∠PAQ = 30° and ∠PBQ = 60°
Let BQ = x and PQ = h
In ΔPBQ,
` tan 60° = (PQ)/(BQ) `
`⇒ sqrt(3) = h/x`
`⇒ = x sqrt(3) ` .......(1)
Also, in ΔAPQ ,
` tan 30° = (PQ)/(AQ)`
`⇒ 1/ sqrt(3) = h/(AB +BQ)`
`⇒ 1/ sqrt(3) = (x sqrt(3))/(20+x)` [Using (1)]
⇒ 20+ x = 3x
⇒ 3x -x = 20
⇒ 2x = 20
`⇒ =20/2`
⇒ x = 10 m
From (i),
`h = 10 sqrt(3)=10xx1.732 =17.32m`
Also, AQ = AB+BQ = 20 + 10 = 30m
So, the height of the tower is 17. 32 m and its distance from the point A is 30 m.
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