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 √3 = 1.732]

In the given figure, let AB be the tower. Let D be the point on the same level as the foot of the tower from which the angle of elevation of the top of the tower is 30°. Join B,C and A,D. Then we get a right-angled triangle ABD with right angle at B and ∠ADB = 30°. Let C be the point on the same level of the ground as Band D, 150 m from D towards B, from which the angle of elevation of the top of the tower is 60°. Joining A and C, we get a right angled triangle ABC, with right angle at B and ∠ACB = 60°. We are to show that the height of the tower AB is 129.9 m.

In ∆ABC,

or,

In ∆ABD,

or,

So, BC = 75m. Now, AB = BC√3 = 75 × 1.732m = 129.9m. Hence proved.