Two stations due south of a tower, which leans towards north are at distances 'a' and 'b' from its foot. If α and β be the elevations of the top of the tower from the situation, Prove that its inclination 'θ' to the horizontal given by

cotΘ =  (bcot α - α cot β) / b-a

Ans:    Let AB be the leaning tower and C and D be the given stations. Draw BL ⊥ DA produced.

Then, ∠BAL = 0, ∠BCA = α, ∠BDC = a and DA = b.

Let AL = x and BL = h

In right ΔALB, we have :

AL/BL = Cot θ ⇒ x/h = Cot θ

⇒  x/h = Cot θ ⇒ x = h cot θ             .....(i)

In right ΔBCL, we have :

CL/BL = Cot α ⇒ a + x = h cot α

⇒ a = h (cot α - cot θ)

⇒ h =a / (cot α - cot θ)             ...(ii)

In right ΔBDL, we have :

DL/BL = cot β ⇒ DA + AL/BL = cot β

⇒ b + x/h = cot β ⇒ b + x = b cot β

⇒ b = h ((cot β - cot θ)           [using (i)]

⇒ h = b/(cot β - cotθ )       ...........(iii)

equating the value of h in (ii) and (iii), we get:

a/(cot α - cot θ )= b /(cot β - cotθ )

⇒ a cot β - a cot θ = b cot α - b cot θ

⇒ (b - a) cot θ = b cot α - a cotβ

⇒ cot θ = b cot α - a cot βθ/(b - a)