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 20m 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.

Asked by Pragya Singh | 1 year ago | 80

Given, AB is the height of the tower.

DC = 20 m (given)

As per given diagram, In right ΔABD,

tan 30° = \( \dfrac{AB}{BD}\)

\(\dfrac{1}{\sqrt{3}} = \dfrac{AB}{20+BC} \)

\(AB+ (20+BC)\sqrt{3}\) … (i)

Again,

In right ΔABC,

tan 60° = \( \dfrac{AB}{BC}\)

\( \sqrt{3}=\) \( \dfrac{AB}{BC}\)

\(AB= \sqrt{3}BC\) … (ii)

From equation (i) and (ii)

\( \sqrt{3}BC=\dfrac{20+BC}{\sqrt{3}} \)

3 BC = 20 + BC

2 BC = 20

BC = 10

Putting the value of BC in equation (ii)

\( AB = 10\sqrt{3}\)

This implies, the height of the tower is \( 10\sqrt{3}\;m\) and the width of the canal is 10 m.

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