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Question 15
A ladder against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p, so that is upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show that pq=cos βcos αsin αsin β

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Solution

Let OQ = x and OA = y

Given that, BQ = q, SA = p and AB = SQ = length of ladder

Also, BAQ=α and QSO=β

Now in ΔBAO,

cos α=OAAB

cos α=yAB

y=AB cos α=OA(i)

And sin α=OBAB

OB=BA sin α

Now, in ΔQSO,

cos β=OSSQ

OS=SQ cos β=AB cos β[AB=SQ](iii)

And sin β=OQSQ

OQ=SQ sin β=AB sin β [AB=SQ](iv)

Now, SA = OS - AO

P=AB cos βAB cos α

P=AB(cos βcos α)(v)

And BQ = BO - QO

q=BA sin αAB sin β

q=AB(sin αsin β)(vi)

Eq. (v) divided by Eq. (vi), we get

pq=AB(cos βcos α)AB(sin αsin β)=cos βcos αsin αsin β

pq=cos βcos αsin αsin β

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