Question

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.

Answer 1

Let the height of the tower is = h (m)

Distance of point B from foot of the tower is = x (m) 

In ∆ADC, 

tan 30° = \(\frac{DC}{AC}\)

\(\frac{1}{\sqrt3}\) = \(\frac{h}{20+x}\)

√3 h = 20+x -----------(1) 

In ∆DCB, 

tan 60° = \(\frac{DC}{BC}\)

√3 = \(\frac{h}{x}\)

h = √3 x----------(2) 

On substituting value of h from eqn. (2) in eqn. (1) 

√3× √3 x = 20 + x 

3 x =  20 + x

3 x - x = 20 

x = 10 

Therefore distance of point A from tower is 

AC = AB + BC 

AC = 20 +10 ⇒ 30 

Ac = 30 m. 

Now substituting value of x in eqn. (1) 

√3 h = 20 +10 ⇒ 30 

h = \(\frac{30}{\sqrt3}\)⇒ 17.32 m. 

Therefore height of tower is 17.32 m.