Question

A ladder rests against a wall at 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 a/b=(cos alpha -cos beta) /(sin beta - sin alpha)

Answer 1

here is the answer by which u can solve Ver easily.

Answer 2

From the given we can prove that a/b=(cos alpha -cos beta) /(sin beta - sin alpha).

Given

• when the foot of the ladder is pulled away from the wall the distance 'a'.

• when the distance slides the distance 'b' it will make

$\angle \: \beta \: horizontal$

Assume the height of the ladder will be BE = AD = h

and AB and CD represent the ladder.

$= > \sin \: \alpha = \frac{ac}{ad} \\ \sin \: \alpha \: h = ac$

$= > \cos \: \alpha \: = \frac{cd}{ad} \\ \cos \alpha \: h = cd$

$= > \sin \: \beta = \frac{bc}{be} \\ \sin \: \beta \: h = bc$

$= > \cos \: \beta = \frac{ce}{be} \\ \cos \: \beta \: h \: = ce$

a = CE - CD

b = AC - BC

Hence from the above we can state that

$\frac{a}{b} = \frac{ce - cd}{ac - bc}$

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

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

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

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

Hence proved.