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Question
A ladder of length 6meters makes an angle of 45° with the floor while leaning against one wall of a room. If the fort 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 floor. Find the
distance between two walls of the room.
Solution
Let AB and CD be the two opposite walls of the room and the foot of the ladder be fixed at
the point O on the ground.
We have,
AO = CO = 6m,∠AOB = 60° and ∠COD = 45°
In ΔABO,
`cos 60°= (BD)/(AO)`
`⇒1/2 = (BO)/6`
`⇒ BO = 6/2`
⇒ BO = 3m
Also, in ΔCDO,
`cos 45° = (DO)/(CO)`
`⇒1/sqrt(2) = (DO)/6`
`⇒ DO = 6/sqrt(2) xx sqrt(2)/sqrt(2)`
`⇒DO = (6 sqrt(2))/2`
`⇒ Do = 3 sqrt(2) m`
Now, the distance between two walls of the room = BD
=BO+DO
`=3+3sqrt(2) `
`=3(1+sqrt(2))`
=3(1+1.414)
=3(2.414)
=7.242
`~~7.24m`
So, the distant between two walls of the room is 7. 24 m.
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