In the above figure, AD and BC are the two opposite walls of the same room. The ladder is fixed at the position say P on the floor. Join D,P and C. At first let it leans against the wall BC making an angle of 45° with the floor. Join B and P. We get a right-angled triangle BPC with right angle at C and ∠BPC = 45°. Again, when the ladder leans against the second wall AD, it makes an angle of 60° with the floor. Joining A and P, we get a right-angled triangle APD with right angle at D and ∠APD = 60°. We are to find the distance between the two walls, that is DC. Also given that, AP = PB = length of the ladder = 6 m.

To find DC, we separately find DP and PC from triangles APD and BPC respectively by using the trigonometric ratio cosine.

From ∆APD,

or,

Again, from ∆BPC,

or,

Hence, the distance between the two walls is, DC = DP + PC = 3 + 4.23 = 7.23m.