From the top of a tower h m high, angles of depression of two objects, which are in line with the foot of the tower are `alpha` and `beta(betagtalpha)`. Find the distance between the two objects.

Explanation

Let the distance betweeen two objects is x m. and CD=y m.
Given that, `angleBAX=alpha=angleABD`, [alternate angles]
`angleCAY=beta= angleACD` (alternate angles)
and the height of the tower, AD=h m
Now, in `DeltaACD`,
`tanbeta=(AD)/(CD) = h/y`
`rArr y=h/(tanbeta)`..............(i)

and in `DeltaABD`,
`tanalpha = (AD)/(BD) rArr = (AD)/(BC+CD)`
`rArr tanalpha=h/(x+y) rArr x+y=h/(tanalpha)`
`rArr y=h/(tanalpha) -x`
From Eqs. (i) and (ii),
`h/(tanbeta) = h/(tanalpha)-x`
`x=h/(tanalpha) - h/(tanbeta)`
`=h(1/(tanalpha)-1/(tanbeta)) = h(cotalpha-cotbeta)` `[therefore cot theta = 1/(tantheta)]`
Which is required distance between the two objects. Hence Proved.