XAT 2014 Question 56

Question 56

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

Solution

Given : Length of ladder = PA = P'B = 30 m and PQ = 26 m

$$\angle$$ PAQ = 2 $$\angle$$ P'BQ

To find : AB = ?

Solution : In $$\triangle$$ PAQ

=> $$(QA)^2 = (PA)^2 - (PQ)^2$$

=> $$(QA)^2 = 30^2 - 26^2 = 900 - 676$$

=> $$QA = \sqrt{224} \approx 15$$

Also, $$cos 2 \theta = \frac{QA}{PA}$$

=> $$2 \theta = cos^{-1}(\frac{15}{30})$$

=> $$2 \theta = 60$$ => $$\theta = \frac{60}{2} = 30$$

In $$\triangle$$ P'QB

=> $$cos 30 = \frac{QB}{P'B}$$

=> $$QB = \frac{\sqrt{3}}{2} \times 30$$

=> $$QB = 15 \times 1.732 = 25.98$$

$$\therefore$$ AB = QB - QA = 25.98 - 15 = 10.98 m



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