A ladder rests against a wall at an angle $ \alpha $ to the horizontal. Its foot is pulled away from the wall through a distance a , so that it slides a distance b down the wall making an angle $ \beta $ with the horizontal. Show that\[\frac{a}{b}=\frac{\cos \alpha-\cos \beta}{\sin \beta-\sin \alpha}\].

Given:

A ladder rests against a wall at an angle \( \alpha \) to the horizontal. Its foot is pulled away from the wall through a distance $a$, so that it slides a distance $b$  down the wall making an angle \( \beta \) with the horizontal.

To do:

We have to show that\[\frac{a}{b}=\frac{\cos \alpha-\cos \beta}{\sin \beta-\sin \alpha}\].

Solution:

From the figure,

$AB$ and $CD$ is the same stair. This implies $AB = CD$

$\cos \alpha=\frac{\text { Base }}{\text { Hypotenuse }}$

$=\frac{\mathrm{AE}}{\mathrm{AB}}$

Similarly,

$\cos \beta=\frac{\mathrm{CE}}{\mathrm{CD}}$

$=\frac{a+\mathrm{AE}}{\mathrm{AB}}$

$\sin \alpha=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$=\frac{\mathrm{BE}}{\mathrm{AB}}$

$=\frac{b+\mathrm{DE}}{\mathrm{AB}}$

$\sin \beta=\frac{\mathrm{DE}}{\mathrm{CD}}$

$=\frac{\mathrm{DE}}{\mathrm{AB}}$

Let us consider RHS,

$\frac{\cos \alpha-\cos \beta}{\sin \beta-\sin \alpha}=\frac{\frac{\mathrm{AE}}{\mathrm{AB}}-\frac{a+\mathrm{AE}}{\mathrm{AB}}}{\frac{\mathrm{DE}}{\mathrm{AB}}-\frac{b+\mathrm{DE}}{\mathrm{AB}}}$

$=\frac{\mathrm{AE}-a-\mathrm{AE}}{\mathrm{DE}-b-\mathrm{DE}}$

$=\frac{-a}{-b}$

$=\frac{a}{b}$

$=$ LHS

Hence proved.

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Updated on: 10-Oct-2022

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