Q. 154.0( 11 Votes )
A TV tower stands vertically on a bank of the canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
In the figure, let AB be the tower and BC be the canal. C is the point on the other side of the canal directly opposite the tower. Join B and C. Let D be another point, 20m away from C. Join C and D. We get two right-angled triangles ∆ABC and ∆ABD, both right angled at B. Given that, ∠ACB = 60° and ∠ADB = 30°. DC = 20m. We use trigonometric ratio tan using AB as height and BC as a base(for ∆ABC) and AB as height and BD as a base(for ∆ABD) to find AB and BC.
Equating the values of AB in ∆ABC and ∆ABD,
So, the width of the canal = BC = 10m. We found that AB = BC√3 = 10√3m.
So, height of the tower = AB =
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