Question

THE ANGLE OF ELEVATION OF THE TOP OF A TOWER AT A DISTANCE OF 100 NETRES FROM ITS FOOT ON A HORIZONTAL PLANE IS FOUND TO BE 60°.FIND THE HEIGHT OF THE TOWER​

Answer 1

Answer:

angle of elevation=60

distance=100m

1. tan60=x/100
2. 1.732=x/100
3. 1.732×100=x
4. x=173.2m