Question

The angle of elevation of the top of a tower from a point A on the ground is 30°. Moving a distance of 20metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower & the distance of the tower from the point A.

Answer 1

Let h be the height of the tower and the angle of elevation of the top of the tower from a point A on the ground is  30° and on moving with distance 20 m  towards the foot of tower on the point B is 60°.

Let AB = 20 and BC = x

Now we have to find the height of tower and distance of tower from point A.

So we use trigonometrical ratios.

In ΔDBC

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

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

`=> sqrt3 = h/x`

`=> x = h/sqrt3`

Again in Δ DAC

`=> tan A = (CD)/(BC + BA)`

`=> tan 30^@ = h/(x + 20)`

`=> 1/sqrt3 = h/(x + 20)`

`=> x = sqrt3h - 20`

`=> h/sqrt3 + 20 = sqrt3h`

`=> h/sqrt3 - sqrt3h = -20`

`=> h - 3h = -20sqrt3`

`=> -2h = -20sqrt3`

`=> h = 10sqrt3`

`=> h = 17.32`

`=> x = (10sqrt3)/sqrt3`

`=> x = 10`

So distance

`=> AC = x + 20`

=> AC = 30

Hence the required height is 17.32 m and distance is 30 m