Question

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.

Answer it Step by Step please​

Answer 1

Step-by-step explanation:

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.