Question

A TV tower stands vertically on a bank
of a 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 joing
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

Answer:

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)

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.

Answer 2

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