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Ladder Against Building Windows

XAT 2014 · Quantitative Ability & Data Interpretation · Hard · Trigonometry

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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

  1. A.

    < 9 m

  2. B.

    ≥ 9 m and < 9.5 m

  3. C.

    ≥ 9.5 m and < 10 m

  4. D.

    ≥ 10 m and < 10.5 m

  5. E.

    ≥ 10.5 m

Answer

E

Explanation

QA=302−262=224≈14.97QA = \sqrt{30^2 - 26^2} = \sqrt{224} \approx 14.97 m. Let ∠PAQ=2θ\angle PAQ = 2\theta, so cos⁡2θ=14.97/30≈0.5  ⟹  2θ≈60∘  ⟹  θ≈30∘\cos 2\theta = 14.97/30 \approx 0.5 \implies 2\theta \approx 60^\circ \implies \theta \approx 30^\circ. QB=30⋅cos⁡30∘=30⋅0.866=25.98QB = 30 \cdot \cos 30^\circ = 30 \cdot 0.866 = 25.98 m. Distance AB=QB−QA=25.98−14.97=11.01AB = QB - QA = 25.98 - 14.97 = 11.01 m ≥10.5\ge 10.5 m.

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