Question

A man observes the angle of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.

Answer 1

Let AB be a building and M and N are the two positions of the man which makes angle of elevation of top of buildings as 30° and 60° respectively. 

MN = 60 m 

Let AB = h and NB = x m 

Now in right ΔAMB, 

`tan 30^circ = (AB)/(MB)`

`=> tan 30^circ = h/(60 + x)`

`=> 1/sqrt(3) = h/(60 + x)`

`=> 60 + x = sqrt(3)h`

`=> x = sqrt(3)h - 60`  ...(1)

Similarly in right ΔANB, 

`tan 60^circ = (AB)/(NB)`

`tan 60^circ = h/(60 + x)`

`=> sqrt(3) = h/x`

`=> x = h/sqrt(3)`  ...(2)

From (1) and (2), we have, 

`sqrt(3)h - 60 = h/sqrt(3)`

`=> 3h - 60sqrt(3) = h`

`=> 3h - h = 60sqrt(3)`

`=> 2h = 60sqrt(3)`

`=> h= (60sqrt(3))/(2)`

`=> h = 30sqrt(3) = 30 xx 1.732`

`=>` h = 51.96 m

∴ Height of the building = 51.96 = 52 m (approx)