A ladder of length 6 metres makes an angle of `45^(@)` with the floor while leaning against one wall of a room. If the foot of the ladder is kept fixed on the floor and it is made to lean against the opposite wall of the room, it makes an angle of `60^(@)` with the floot. Find the distance between two walls of the room.

Answer

`7.24 m`

Explanation

Let AB and CD be the two opposite walls and let the foot of the ladder be fixed point O on the ground. Let OB and OD be the two positions of the ladder.
From right `DeltaOAB`, we have
`(OA)/(OB) = cos 45^(@) rArr (OA)/(6m) = (1)/(sqrt(2))`.
`rArr OA = (6 m)/(sqrt(2)) rArr OA = (6 m)/(sqrt(2)) xx (sqrt(2))/(sqrt(2)) = 3sqrt(2) m`.
From right `DeltaOCD`, we have
`(OC)/(OD) =cos 60^(@) rArr (OC)/(6 m) = 1/2 rArr OC = 3 m`.
Distance between two walls `= AC = OA + OC = 3(sqrt(2) + 1) m = 3(1.414+1)`
`= (3 xx 2.414) m = 7.242 m = 7.24 m`.