A ladder rests against a wall at an angle α to the horizontal.

Solution:

Let be the ladder such that its top  is on the wall  and bottom  is on the ground. The ladder is pulled away from the wall through a distance a, so that its top  slides and takes position. So

$\angle O P Q=\alpha$ And $\angle O P^{\prime} Q^{\prime}=\beta$. Let $P Q=h$

We have to prove that

$\frac{a}{b}=\frac{\cos \alpha-\cos \beta}{\sin \beta-\sin \alpha}$

We have the corresponding figure as follows

We use trigonometric ratios.

In $\triangle P O Q$

$\Rightarrow \quad \sin \alpha=\frac{O Q}{P O}$

$\Rightarrow \quad \sin \alpha=\frac{b+y}{h}$

And

$\Rightarrow \quad \cos \alpha=\frac{O P}{P Q}$

$\Rightarrow \quad \cos \alpha=\frac{x}{h}$

Again in $\triangle P^{\prime} O Q^{\prime}$

$\Rightarrow \quad \sin \beta=\frac{O Q^{\prime}}{P^{\prime} Q^{\prime}}$

$\Rightarrow \quad \sin \beta=\frac{y}{h}$

And

$\Rightarrow \quad \cos \beta=\frac{O P^{\prime}}{P^{\prime} O^{\prime}}$

$\Rightarrow \quad \cos \beta=\frac{a+x}{h}$

Now,

$\Rightarrow \sin \alpha-\sin \beta=\frac{b+y}{h}-\frac{y}{h}$

$\Rightarrow \sin \alpha-\sin \beta=\frac{b}{h}$

So

$\Rightarrow \frac{\sin \alpha-\sin \beta}{\cos \beta-\cos \alpha}=\frac{b}{a}$

$\Rightarrow \quad \frac{a}{b}=\frac{\cos \beta-\cos \alpha}{\sin \alpha-\sin \beta}$

$\Rightarrow \quad \frac{a}{b}=\frac{\cos \alpha-\cos \beta}{\sin \beta-\sin \alpha}$

Hence $\frac{a}{b}=\frac{\cos \alpha-\cos \beta}{\sin \beta-\sin \alpha}$.