Question

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

Answer 1

To show $\frac{a}{b}=\frac{cos\alpha -cos\beta }{sin\beta -sin\alpha }\dots \dots \left(1\right)$
$\Rightarrow cos\alpha =\frac{BD}{DE}=\frac{AB-a}{DE}$
$\Rightarrow cos\beta =\frac{AB}{AC}$
$\Rightarrow sin\beta =\frac{BC}{AC}=\frac{EB-b}{AC}$
$\Rightarrow sin\alpha =\frac{EB}{DE}$
Put in (1)

$\Rightarrow \frac{cos\alpha -cos\beta }{sin\beta -sin\alpha }=\frac{\frac{AB-a}{DE}-\frac{AB}{AC}}{\frac{EB-b}{AC}-\frac{EB}{DE}}$
DE=AC=Length of ladder
$\Rightarrow \frac{cos\alpha -cos\beta }{sin\beta -sin\alpha }=\frac{\frac{AB-a}{AC}-\frac{AB}{AC}}{\frac{EB-b}{AC}-\frac{EB}{AC}}$
$\Rightarrow \frac{\frac{AB-a-AB}{AC}}{\frac{EB-b-EB}{AC}}$
$=\frac{-a}{-b}=\frac{a}{b}$