Question

from a point A the angle of elevation of the top of vertical tower situated on the roof of 50m high building is found to be theta .after walking some distance towards the tower the angle of elevation of bottom of tower from point B is also theta.find the height of tower.

Answer 1

From ΔABC:

tan (θ) = opp/adj

tan (30) = BC/AB

BC = AB tan (30)

From ΔBCD:

tan (θ) = opp/adj

tan (60) = BC/BD

BC = BD tan (60)

Equate the 2 equations:

AB tan (60) = BD tan (30)

Define x:

Let BD = x

AB = x + 20

Solve x:

AB tan (30) = BD tan (60)

(x + 20) tan (30) = x tan (60)

x tan (30) + 20 tan (30) = x tan (60)

x tan (60) - x tan (30) = 20 tan (30)

x ( tan (60) - tan (30) ) = 20 tan (30)

x = 20 tan (30) ÷ ( tan (60) - tan (60) )

x = 10 m

Find the distance:

Distance = 10 + 20 = 30 m

Find the height:

tan (θ) = opp/adj

tan (60) = BC/10

BC = 10 tan (60)  = 10√3 m

Answer:  Distance = 30 m and height = 10√3 m

Answer 2

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