A ladder rests against a wall making an angle | vedclass

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Question

(A) Let $l$ be the length of the ladder.In the initial position,the ladder makes an angle $\alpha$ with the horizontal. The horizontal distance from t

Vedclass · Accepted Answer

$x = y 	an \frac{\alpha + \beta}{2}$

Vedclass · Answer

(A) Let $l$ be the length of the ladder.In the initial position,the ladder makes an angle $\alpha$ with the horizontal. The horizontal distance from the wall is $PA = l \cos \alpha$ and the vertical height is $AB = l \sin \alpha$.In the final position,the ladder makes an angle $\beta$ with the horizontal. The horizontal distance from the wall is $QA = l \cos \beta$ and the vertical height is $AC = l \sin \beta$.The foot of the ladder is pulled away by distance $x = QA - PA = l(\cos \beta - \cos \alpha)$.The ladder slides down by distance $y = AB - AC = l(\sin \alpha - \sin \beta)$.Now,taking the ratio $\frac{y}{x}$:$\frac{y}{x} = \frac{l(\sin \alpha - \sin \beta)}{l(\cos \beta - \cos \alpha)} = \frac{2 \sin \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2}}{2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}}$$\frac{y}{x} = \cot \frac{\alpha + \beta}{2}$Therefore,$x = y 	an \frac{\alpha + \beta}{2}$.

$A$ ladder rests against a wall making an angle $\alpha$ with the horizontal. The foot of the ladder is pulled away from the wall through a distance $x$,so that it slides a distance $y$ down the wall making an angle $\beta$ with the horizontal. The correct relation is

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