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.

Let AB is TV tower and BC is width of the canal. In ΔABC,
AB/BC = tan 60° ⇒ AB/BC = √3
⇒ BC = AB√3
In ΔABD,
AB/BD = tan 30° ⇒ AB/(BC+CD) = 1/√3
⇒ AB/(AB/√3+CD) = 1/√3 [Putting the value of BC]
⇒ AB√3/(AB + 20√3) = 1/√3 ⇒ 3AB = AB + 20√3 ⇒ 2AB = 20√3 ⇒AB = 10√3
Therefore, BC = AB/√3 = 10√3/√3= 10
Hence, the height of the tower is 10√3m and the width of tower is 10m.