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Question

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)#

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Answer 1 · 2018-06-05T02:12:07

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Answer 2 · 2018-06-05T02:48:02

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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