Question

A TV tower stands vertically on a bank of canal. From a point on 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.

Answer 1

In the figure, let AB be the tower and AC be the canal. C is the point on the other side of the canal directly opposite the tower.

Let AB = h, CA = x and CD = 20

From right △BAC:

cot 60° = x/h

1/√3 = x/h

x = h/√3 …(1)

From right △BAD:

cot 30° = AD/AB

√3 = (x+20)/h

x = h√3 – 20 …(2)

From (1) and (2), we have

h/√3 = h√3 – 20

h = 10√3

From (1): x = (10√3)/√3 = 10

Therefore, the width of the canal is 10 m and height of the canal is 10√3 m.