Q. 214.3( 6 Votes )
The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower.
In the figure, let AB be the tower and C be the point on the ground from which the angle of elevation of the top of the tower AB is 60°. Join A and C and B and C. Then we get a right-angled triangle ABC with right angle at B and ∠ACB = 60°. Let D be the point 10 m vertically above C. Then, CD = 10 m. Given that the angle of elevation of the top of the tower from the point D is 30°. Join D and A. Also draw a line DE from D to AB, parallel to BC. Then we get a right-angled triangle AED with right angle at E and ∠ADE = 30°. We are to find the height of the tower, that is AB.
Clearly, ED = BC. Let AE = x
Again, in ∆ABC,
3x = x + 10"
x = 5m
Height of the tower = AB = AE + EB = 5m + 10m = 15m.
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