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
CD2 = 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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