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.