Question

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

Answer 1

The given information can be represented diagrammatically as

In the above figure $AB$ and $CD$ represent the same ladder
So, their length must be equal
Let length of the ladder be h
∴ AB = CD = h

$In△AEB:\frac{AE}{AB}=\sin \alpha \text{and}\frac{BE}{AB}=\cos \alpha$

=> $AE=h\sin \alpha \text{and}BE=h\cos \alpha$

$In△DEC:\frac{DE}{CD}=\sin \beta \text{and}\frac{CE}{CD}=\cos \beta$

=> $DE=h\sin \beta ,CE=h\cos \beta$

$Now,\frac{p}{q}=\frac{BC}{AD}=\frac{CE-BE}{AEC-DE}$

$\Rightarrow \frac{hcos\beta -hcos\alpha }{hsin\alpha -hsin\beta }$

$\Rightarrow \frac{p}{q}=\frac{cos\beta -cos\alpha }{sin\alpha -sin\beta }$

Hence, proved