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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

Explanation:

The following figure represents the length and the position of the ladder,

By Pythagoras Theorem,

$x = \sqrt{30^{2} - 26^{2}} = \sqrt{224} \approx 15 m$

$∴ \cos 2 θ = \frac{15}{30} = \frac{1}{2}$

$\Rightarrow 2 θ = 60 ° \Rightarrow θ = 30 °$

$\cos 30 ° = \frac{y}{30}$

$\Rightarrow \frac{\sqrt{3}}{2} = \frac{y}{30}$

$\Rightarrow y = 15 \sqrt{3} m$

The approximate distance between A and B can be given as,

$y - x = 15 \sqrt{3} - 15 = 10.98 m$

Hence, option (e)

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