Question

The angle of elevation of the top of a tower, from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.

Answer 1

Let AB be the tower and C is the point which is 160 m away from the foot of the tower, i.e CB = 160 m

Let height of the tower be h

Now in right ΔABC, we have

`tan theta = (AB)/(BC)`

`=> tan 60^circ = h/160`

∴ h = 160 × tan 60°

= `160 xx sqrt(3)  m`

= 160 × (1.732) m

= 277.12 m

∴ Required height of the tower = 277.12 m