Question

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

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 m towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and 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^circ = (CD)/(BC)`

⇒ `sqrt(3) = h/x`

⇒ `x = h/sqrt(3)`

Again in ΔDAC

⇒ `tan A = (CD)/(BC + BA)`

⇒ `tan 30^circ = h/(x + 20)`

⇒ `1/sqrt(3) = h/(x + 20)`

⇒ `x = sqrt(3)h - 20`

⇒ `h/sqrt(3) + 20 = sqrt(3)h`

⇒ `h/sqrt3 - sqrt3h = -20`

⇒ `h - 3h = -20sqrt(3)`

⇒ `-2h = -20sqrt(3)`

⇒ `h = 10sqrt(3)`

⇒ h = 17.32

⇒ `x = (10sqrt(3))/sqrt(3)`

⇒ x = 10

So distance

⇒ AC = x + 20

⇒ AC = 30

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