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A ladder rests against a vertical wall a...

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)`.

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Let a ladder AB of lengt 'l', resting against the wall such that `angleABE=alpha`
when its foot pulled away from the wall at a distance 'p' then new position of the ladder becomes CD.
Now, `CD=AB=l`
Given that, `angleDCE=beta" and "AD=q`
Let `BE=x" and "DE=y`
In `DeltaABE`,
`sinalpha=(AE)/(AB)=(q+y)/1" and "cosalpha=(BE)/(AB)=x/l`
In `DeltaCDE`,
`sinbeta=(DE)/(CD)=y/1" and "cosbeta=(CE)/(CD)=(p+x)/l`
Now, `cos beta-cos alpha=(p+x)/l-x/l=p/l ...(1)`
and `sinalpha-ainbeta=(q+y)/l-x/l=q/1 ...(2)`
Divide equation (1) by equation (2), we get
`(cos beta-cosalpha)/(sinalpha-sinbeta)=(p//l)/(q//l)=p/q`
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