Question

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 floor. Find the distance between two walls of the room. [CBSE 2011]

Answer 1

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,

$$\begin{gathered} \mathrm{AO}=\mathrm{CO}=6\mathrm{m},\angle \mathrm{AOB}=60°\mathrm{and}\angle \mathrm{COD}=45° \\ \mathrm{In}∆\mathrm{ABO}, \\ \cos 60°=\frac{\mathrm{BO}}{\mathrm{AO}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{BO}}{6} \\ \Rightarrow \mathrm{BO}=\frac{6}{2} \\ \Rightarrow \mathrm{BO}=3\mathrm{m} \\ \mathrm{Also},\mathrm{in}∆\mathrm{CDO}, \\ \cos 45°=\frac{\mathrm{DO}}{\mathrm{CO}} \\ \Rightarrow \frac{1}{\sqrt{2}}=\frac{\mathrm{DO}}{6} \\ \Rightarrow \mathrm{DO}=\frac{6}{\sqrt{2}} \\ \Rightarrow \mathrm{DO}=\frac{6}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ \Rightarrow \mathrm{DO}=\frac{6\sqrt{2}}{2} \\ \Rightarrow \mathrm{DO}=3\sqrt{2}\mathrm{m} \\ \mathrm{Now},\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{two}\mathrm{walls}\mathrm{of}\mathrm{the}\mathrm{room}=\mathrm{BD} \\ =\mathrm{BO}+\mathrm{DO} \\ =3+3\sqrt{2} \\ =3\left(1+\sqrt{2}\right) \\ =3\left(1+1.414\right) \\ =3\left(2.414\right) \\ =7.242 \\ \approx 7.24\mathrm{m} \end{gathered}$$

So, the distance between two walls of the room is 7.24 m.