Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot If `alpha` and `beta` are the elevations of the top of the tower from these stations then prove that its inclination `theta` to the horizontal is given by `cottheta =(bcotalpha-acotbeta)/(b-a)`

Explanation

Let AB be a leaning tower and let C and D be two stations due south at distances a and b respectively from the foot A of the tower.
Let AE = x and BE = h
In right `DeltaBEA`,
`tantheta=h/xrArrx=h cottheta …(1)`
In `DeltaCEB`,
`tanalphah/(a+x)rArr a+x=h cot alpha ...(2)`
In `DeltaDEB`,
`tanbeta=h/(b+x)rArr b+x=h cotbeta ...(3)`
Multiplying equation (2) by b and equation (3) by a, we get
`ab+bx=bhcotalpha ...(4)`
and `(_abpmax=pmahcotbeta)/(x(b-a)=h(bcotalpha-alphacotbeta))`
`:. x/h=(b cotalpha-a cot beta)/(b-a)rArrcottheta=(b cotalpha-acotbeta)/(b-a)" [from(1)]"`