A ladder rests against a vertical wall at inclination `alpha` to the horizontal. Its foot is pulled away from the wall through a distance p so that it's upper end slides q down the wall and then ladder make an angle `beta` to the horizontal show that ` p/q = ( cos beta - cos alpha)/(sin alpha - sin beta)`.

Explanation

Let OQ=x and OA=y
Given that, BQ=q, SA=P and AB=SQ=Length of ladder
Also, `angleBAO = alpha` and `angleQSO=beta`
Now, in `DeltaQSO`
`cosbeta=(OS)/(SQ)`
`rArr OS = Sqcosbeta=Abcosbeta` `[therefore AB=SQ]`………….(iii)
and `sinbeta=(OQ)/(SQ)`
`rArr OQ=SQ sin beta-ABsinbeta` `[therefore AB=SQ]`...............(iv)

Now, SA=OS-AO
`P=ABcosbeta-ABcosalpha`
`rArr P=AB(cosbeta-cosalpha)`............(v)
and BQ=BO-QO
`rArr q=BA(sinalpha-sinbeta)`..............(vi)
Eq. (v) divided by Eq.(vi), we get
`p/q= (AB(cosbeta-cosalpha))/(AB(sinalpha-sinbeta)) = (cosbeta-cosalpha)/(sinalpha-sinbeta)`
`rArr p/q = (cosbeta-cosalpha)/(sinalpha-sinbeta)` Hence proved.