Question

The question is below?

A ladder rests against a wall with an angle `alpha` to the horizontal. Its foot is being pulled away from the wall through a distance `a` so that it slides `a` distance `b` down the wall making an angle `beta` with the horizontal. Show that `a=b*tan((alpha+beta)/2)`

Answer 1

Duplicate deleted

Answer 2

Let AB and CD be 1st and 2nd positions of the ladder. So `AB=CD =x (say)`.
Again as per problem `BC=a and AD=b`

Now

`a=BC=EC-EB=((EC)/(CD)-(EB)/(CD))*CD`

`=>a=((EC)/(CD)-(EB)/(AB))*CD`

`=>a=(cosbeta-cosalpha)x...[1]`

Similarly

`b=AD=AE-ED=((AE)/(AB)-(ED)/(AB))*AB`

`=>b=((AE)/(AB)-(ED)/(CD))*AB`

`=>b=(sinalpha-sinbeta)x...[2]`

Dividing [1] by [2] we get

`a/b=(cosbeta-cosalpha)/(sinalpha-sinbeta)`

`=>a/b=(2sin((alpha+beta)/2)sin((alpha-beta)/2))/(2cos((alpha+beta)/2)sin((alpha-beta)/2))`

Hence

`a=btan((alpha+beta)/2)`

Proved