Question

A 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 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.
[CBSE 2009] ​

Answer 1

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°

• Also see the above picture

Answer 2

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Width of canal = 10 m

Height of tower = 17.321 m

Step-by-step explanation:

Let the width of the canal be x.

The total distance from the point 20m away from the opposite bank to the other bank will be : x + 20

We have two right angled triangles with the following properties :

1) Base = x + 20, adjacent angle = 30, height = h

2) Base = x, adjacent angle = 60, height = h

They have a common height.

We will use Tangent to form two equations as follows :

1) Tan 30 = h/(x + 20)

h = (x + 20) Tan 30.............i)

2) Tan 60 = h/x

h = x Tan 60.............ii)

Since in both cases h is equal we equate i and ii.

(x + 20) Tan 30 = x Tan 60

(x + 20) / x = Tan 60 / Tan 30

Tan 60 = 3^½

Tan 30 = (3^½)/3

Substituting we have :

(x + 20) / x = 3^½ × 3/3^½

(x + 20)/x = 3

x + 20 = 3x

3x - x = 20

2x = 20

x = 10

Lets get the value of h now:

h = 10 × 3^½ = 17.321 m

The width of the canal = 10 m

The height of the tower = 17.321 m

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