Question

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is `5/12`. On walking 192 metres towards the tower, the tangent of the angle is found to be `3/4`. Find the height of the tower.

Answer 1

Let AB be the vertical tower and C and D be two point such that CD = 192 m. Let ∠ACB = θ and ∠ADB = α 

Given, `tan theta = 5/12`

`=> (AB)/(BC) = 5/12`

`=> AB = 5/12 BC`    ...(i)

Also, `tan alpha = 3/4`

`=> (AB)/(BD) = 3/4`

`=> ((5)/(12)BC)/(BD) = 3/4`

`=> (192 + BD)/(BD) = 3/4 xx 12/5`

`=>` BD = 240 m 

∴ BC = (192 + 240) = 432 m

∴ By (i), `AB = 5/12 xx 432 = 180  m` 

Hence, the height of the tower is 180 m.