Question

from the top of a building , the angle of elevation of the top of a cell tower is 45 degree and the angle of depression to its foot is 60 degree . if height of cell tower is 46 meters then find the distance between building and tower, and height of the building

Answer 1

tanθ = opp/adj

Define h and d:

Let the height of the building be h

And the distance between the building and the tower be d

From the top of the building to top of the tower:

tanθ = opp/adj

tan(45º) = (46-h)/d

d= (46-h)/tan(45)

From the top of the building to foot of the tower:

tanθ = opp/adj

tan(60º) = (h)/d

d = (h)/tan(60)

Since the distance is the same:

(46-h)/tan(45) = (h)/tan(60)

Since tan(45) = 1 and tan(60) = √3:

46-h= (h)/√3

(46 - h)√3 = h

46√3 - ht√3 = h

h + h√3  = 46√3

h(t1 +√3) = 46√3

h = 46√3  ÷ (1 + √3)

h = 29.16m

Find the distance between the building and the tower:

d = (h)/tan(60)

d = 29.16 / tan(60) = 16.84 m

Answer: The height of the building is 29.16m and the distance between the building and the tower is 16.84m

Answer 2

Answer:

19.1243 m

Step-by-step explanation:

From the top of a building, the angle of elevation of the top of a cell tower is 60° i.e.∠AEB = 60°

The angle of depression to its foot is 45°.i.e. ∠ECB = 45°

Distance of the building from the tower is 7 m i.e. EB =DC = 7m

In ΔAEB

In ΔEDC

So, ED = BC = 7 m

Height of tower = AB +BC =

Hence  the height of the tower is 19.1243 m

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