Question

There are two windows on the wall of a building that need repairs. A ladder 30 m long is placed against a wall such that it just reaches the first window which is 26 m high. The foot of the ladder is at point A. After the first window is fixed, the foot of the ladder is pushed backwards to point B so that the ladder can reach the second window. The angle made by the ladder with the ground is reduced by half, as a result of pushing the ladder. The distance between points A and B is

Accepted Answer

E. $QA = \sqrt{30^2 - 26^2} = \sqrt{224} \approx 14.97$ m. Let $\angle PAQ = 2	heta$, so $\cos 2	heta = 14.97/30 \approx 0.5 \implies 2	heta \approx 60^\circ \implies 	heta \approx 30^\circ$. $QB = 30 \cdot \cos 30^\circ = 30 \cdot 0.866 = 25.98$ m. Distance $AB = QB - QA = 25.98 - 14.97 = 11.01$ m $\ge 10.5$ m.

Ladder Against Building Windows

Explanation

Practise this under exam conditions