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 joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.
Here, AB is the height of the tower.
CD = 20 m (given)
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(20+BC)
⇒ AB = (20+BC)/√3 ... (i)
also,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ AB = √3 BC ... (ii)
From eqn (i) and (ii)
AB = √3 BC = (20+BC)/√3
⇒ 3 BC = 20 + BC
⇒ 2 BC = 20 ⇒ BC = 10 m
Putting the value of BC in eqn (ii)
AB = 10√3 m
Thus, the height of the tower 10√3 m and the width of the canal is 10 m.
