Question

Two stations due south of a leaning tower which leans towards the north are at a distance a and b from its foot .If alpha and beta are the elevation of the top of the tower from these stations then prove that it's inclination theta to the horizontal is given by

Cot theta=bcotalpha - acotbeta/b-a

Answer 1

let height of the tower DE = h
Distance between first station to foot of tower AD = a + x
Distance between second  station to foot of tower BD = b +x
Distance between C and D = x
Give α, β are the angle of elevation two stations to top of the tower
that is ∠DAE = α, ∠DBE = β ,∠DCE =θ .
In Δ ADE
Cot θ = x / h ----------------→(1)
In Δ BDE
Cot β = (b+x) / h
(b+x) = h Cot β         (multiply a on both sides )
(ab+ax) = ha Cot β  ----------------→(2)

In Δ CDE
Cot α = (a+x) / h
(a+x) = h Cot α        (multiply b on both sides )
(ab+bx) = hb Cot α  ----------------→(3)
substract (3) - (2)
(b - a)x = h (b Cot α - aCot β)
x / h = (b Cot α - aCot β) / (b - a)
Cot θ = (b Cot α - aCot β) / (b - a).