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).` 

Answer 1

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.