The angle of elevation of the top of a tower from a point A (on the ground) is 30⁰. On walking 50 m towards the tower, the angle of elevation is found to be 60⁰. Calculate

(i) the height of the tower (correct to one decimal place).

(ii) the distance of the tower from A.

Solution:

Consider TR as the tower and A as the point on the ground

Angle of elevation of the top of tower = 30⁰

AB = 50 m

Angle of elevation from B = 60⁰

Take TR = h and AR = x

BR = x – 50

In right triangle ATR

tan θ = TR/AR

Substituting the values

tan 30⁰ = h/x

So we get

1/√3 = h/x

x = √3h ……. (1)

In right triangle BTR

tan θ = TR/BR

Substituting the values

tan 60⁰ = h/ (x – 50)

So we get

√3 = h/ (x – 50)

h = √3 (x – 50) …… (2)

Using both the equations

h = √3 (√3h – 50)

By further calculation

h = 3h – 50√3

2h = 50√3

h = 25 √3

So we get

h = 25 × 1.732 = 43.3

Now substituting the values of h in equation (1)

x = √3 × 25√3

x = 25 × 3

x = 75

Here

Height of the tower = 43.3 m

Distance of A from the foot of the tower = 75 m