Question

The angle of elevation of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 metres towards the foot of the tower, the angle of elevation of the tower becomes 60°.  Show that the height of the tower is 129.9 metres (Use `sqrt3 = 1.732`)

Answer 1

Let h be the height of tower and angle of elevation of the foot of the tower is 30°, on advancing 150 m towards the foot of tower then an angle of elevation becomes 60°.

We assume that BC = x and CD = 150 m.

Now we have to prove height of the tower is 129.9 m.

So we use trigonometrical ratios.

In a triangle ABC

`=> tan C = (AB)/(BC)`

`=> tan 60^@ = (AB)/(BC)`

`=> sqrt3 = h/x`

`=> h/sqrt3 = x`

Again in a triangle ABD

`=> tan D = (AB)/(BC + CD)`

`=> tan 30^@ = h/(x + 150)`

`=> 1/sqrt3 = h/(x + 150)`

`=> x + 150 = sqrt3h`

`=> x = sqrt3h - 150`

`=> h/sqrt3 = sqrt3 - 150`

`=> h = 3h - 150sqrt3`

`=> 2h = 150sqrt3`

`=> h = (150 xx 1.732)/2`

=> h = 129.9

Hence the height of tower is  129.9 m proved