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:
Let ∠ABD = θ then ∠ACB = 90 - θ
Now in Δ ABD
tanθ = AB/BD
= AB/9
Again in Δ ABC
tan(90-θ) = AB/BC
=> cotθ = AB/4
=> 1/tanθ = AB/4
=> tanθ = 4/AB
=> AB/9 = 4/AB (since tanθ = AB/9)
=> AB*AB = 9*4
=> AB2 = 36
=> AB = √36
=> AB = 6
So height of the tower is 6 m.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
