# 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 be CD. B is a point 4 m 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 then the other is (90° - 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 that the height of the tower is 6 m.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 9

**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

**Summary:**

If 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, then the height of the tower is 6 m.

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