# The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m

**Solution:**

Let the height of the tower as CD. B is a point 4m away from the base C of the tower and A is a point 5 m away from point B in the same straight line. The angles of elevation of the top D of the tower from points B and A are complementary.

Since the angles are complementary if one angle is θ and the other is (90° - θ).

The trigonometric ratio involving CD, BC, AC, and angles are tan θ.

Using tan θ and tan (90° - θ) = cot θ ratios are equated to find the height of the tower.

In ΔBCD,

tan θ = CD/BC

tanθ = CD/4 ....(1)

Here, AC = AB + BC = 5 + 4 = 9

In ΔACD,

tan (90 - θ) = CD/AC

cot θ = CD / 9 [Since tan (90- θ) = cot θ]

1 / tanθ = CD / 9 [As we know that cot θ = 1 / tan θ]

tanθ = 9 / CD ....(2)

From equation (1) and (2)

CD / 4 = 9 / CD

CD^{2} = 36

CD = ± 6

Since height cannot be negative, therefore, the height of the tower is 6 m.

Hence proved.

**Video Solution:**

## The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m

### Maths NCERT Solutions Class 10 - Chapter 9 Exercise 9.1 Question 16:

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m

If the angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary, then the height of the tower is 6m