Question

the angle of elevation of the top of a tower from a point on the ground is 45degree on walking 30 metres towards the tower the angle of elevation becomes 60degree find the height of the tower and the original distance from the foot of the tower

Answer 1

Let AB be height of the tower [h ]m

and BC be [x ]m.

According to the question, DB is 30 m longer than BC.

So, DB = (30 + x) m

Now, we have two right triangles ABC and ABD.

In Δ ABC, tan 60° = AB/BC  =perp/ base

or, √3 = h/x ....................(1)

In Δ ABD, tan 30° = AB/BD

or, 1/√3 = h/x ..................(2)

From (1), we have h= x√3

Putting this value in (2), we get (x√3)√3 = x+30, i.e., 3x = x + 30

i.e., x = 15

DB = (30 + x) m = 30+15 =45

original distance from the foot of the tower   =  45 ANS

So, h = 15√3 ..................[From (1)]

Therefore, the height of the tower is   15√3 m. =25.98 m   ANS