# 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° (see Fig. 9.12). Find the height of the tower and the width of the canal.

**Solution:**

Considering ΔABC,

tan 60° = AB/BC

√3 = AB/BC

AB = BC√3 ...(i)

Considering ΔABD,

tan 30° = AB/BD

tan 30° = AB / (CD + BC)

1/√3 = BC√3 / (20 + BC) [from (i)]

20 + BC = BC√3 × √3

20 + BC = 3 × BC

3BC - BC = 20

2BC = 20

BC = 10

Substituting BC = 10 m in Equation (1), we get AB = 10√3 m

Height of the tower AB = 10√3 m

Width of the canal BC = 10 m

**Video Solution:**

## 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° (see Fig. 9.12). Find the height of the tower and the width of the canal.

### Maths NCERT Solutions Class 10 - Chapter 9 Exercise 9.1 Question 11:

If a TV tower stands vertically on the bank of a canal, and from a point on the other bank directly opposite the tower, if the angle of elevation of the top of the tower is 60°, and 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°, then the height of the tower and the width of the canal are 10√3 m and 10 m respectively.