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Question
A ladder rests against a wall at an angle a, to the horizontal. Its foot is pulled away from the wall through a distance 'a', so that it slides a distance 'b' down the wall making an angle β with the horizontal. Show that `"a"/"b" = (cosα - cosβ)/(sinβ - sinα).`
Solution
Let CP and DW be the two position of the ladder such that CP = DW = x (say).
CD = a , PW = b, ∠ACP = α and ∠ADW = β
In ΔAPC,
`"AC"/"CP" = cos α ⇒ "AC" = "x" cosα"` ....(i)
In ΔADW,
`"AD"/"DW" = cos β ⇒ ("AC + CD")/("DW") = cos β`
⇒ `("x" cosα + a)/"x" = cos β` [using (i)]
⇒ `"x" = a/(cos β - cos α)` ...(ii)
Again in ΔAPC, `"AP"/"CP" = sin α`
⇒ `"AP" = "x" sin α = (a sinα)/((cos β - cosα))` ....(iii) [Using (ii)]
Again in ΔADW, `"AW"/"DW" = sin β`
⇒ `"AW" = "x" sin β = (a sin β)/((cos β - cos α))` ...(iv)
Now , PW = AP - AW
⇒ `"b" = ((asinα)/(cosβ -cosα)) - ((asinβ)/(cosβ - cosα)) = (a(sinα -sinβ))/((cosβ - cosα))`
⇒ `a/b = (cosβ - cosα)/(sinα -sinβ)`
Hence proved.
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