Question

A ladder rests against a wall at an angle αα.

Answer 1

Let $BC$ be the ladder which slides down a distance $b$ on the wall.

In right angled $△ABC$, we have

$\sin \alpha =\frac{AB}{BC}=\frac{AE+EB}{BC}$

$\sin \alpha =\frac{AE+b}{BC}$

But, $AE=\sin \beta \times ED$ [ In $△AED$ ]

So, replacing $AE$ by $ED$ $\sin \beta$, we get

$\Rightarrow$ $\sin \alpha =\frac{ED\sin \beta +b}{BC}$

$\Rightarrow$ $b=BC\sin \alpha -ED\sin \beta$

As, $BC$ and $ED$ both represents the same ladder.

$BC=ED$ [ Length of the ladder does not change ]

$\Rightarrow$ $BC\sin \alpha -BC\sin \beta =b$

$\Rightarrow$ $BC(\sin \alpha -\sin \beta )=b$ ------ ( 1 )

Similarly, in right-angled $△AED$, we have

$\cos \beta =\frac{AD}{ED}=\frac{AC+CD}{ED}$

$\cos \beta =\frac{AC+a}{ED}$

But, In $△ABC$, $AC=BC\cos \alpha$

So, by replacing $AC$ by $BC\cos \alpha$, we get,

$cos\beta =\frac{BC\cos \alpha +a}{ED}$

$\therefore$ $BC(\cos \beta -cos\alpha )=a$ ------ ( 2 ) [ $ED=BC$ ]

(From 1 and 2)

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

By arranging we get,

$\Rightarrow$ $\frac{a}{b}=\frac{\cos \alpha -cos\beta }{\sin \beta -sin\alpha }$ [ Hence, proved ]