Question

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres. [Given `sqrt(3) = 1.732`.]

Answer 1

Let AB be the tower
We have:
CD =150m,∠ACB = 30° and ∠ADB = 60°
Let:
AB = hm and BD = xm

In the right ΔABD,we have:

`(AB)/(AD) = tan 60° = sqrt(3)`

`⇒h/x = sqrt(3)`

`⇒ x = h/sqrt(3)`

Now, in the right  ΔACB,we have:

`(AB)/(AC) = tan 30° = 1/ sqrt(3)`

`⇒ h/(x +150) = 1/ sqrt(3)`

`sqrt(3) h =x +150`

On putting `x= h/ sqrt(3) ` in the above equation, we get:

`sqrt(3)h = h/ sqrt(3) + 150`

`⇒ 3h = h + 150 sqrt(3)`

`⇒ 2h = 150sqrt(3)`

`⇒ h = (150 sqrt(3)) /2 =75 sqrt(3) = 75 xx1.732 = 129.9m`

Hence, the height of the tower is 129.9 m