Question

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

Answer 1

Let the height of the building be AB = h and BC = x

In ΔABC

`tan 60^@ = h/x`

`=> xsqrt3 = h`    ...(1)

In ΔADB

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

`=> 1/sqrt3 = h/(x + 60)`

`=> x + 60 = hsqrt3`   ... (2)

From (1)and (2)

`=> x + 60 = xsqrt3.sqrt3`

`=> x= 30`

Thus `h = 3sqrt3 = 51.96m ~~ "52 metre"`