# The question is below?

## A ladder rests against a wall with an angle $\alpha$ to the horizontal. Its foot is being 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 $a = b \cdot \tan \left(\frac{\alpha + \beta}{2}\right)$

Jun 5, 2018

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Jun 5, 2018 Let AB and CD be 1st and 2nd positions of the ladder. So $A B = C D = x \left(s a y\right)$.
Again as per problem $B C = a \mathmr{and} A D = b$

Now

$a = B C = E C - E B = \left(\frac{E C}{C D} - \frac{E B}{C D}\right) \cdot C D$

$\implies a = \left(\frac{E C}{C D} - \frac{E B}{A B}\right) \cdot C D$

$\implies a = \left(\cos \beta - \cos \alpha\right) x \ldots \left[1\right]$

Similarly

$b = A D = A E - E D = \left(\frac{A E}{A B} - \frac{E D}{A B}\right) \cdot A B$

$\implies b = \left(\frac{A E}{A B} - \frac{E D}{C D}\right) \cdot A B$

$\implies b = \left(\sin \alpha - \sin \beta\right) x \ldots \left[2\right]$

Dividing  by  we get

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

$\implies \frac{a}{b} = \frac{2 \sin \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)}{2 \cos \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)}$

Hence

$a = b \tan \left(\frac{\alpha + \beta}{2}\right)$

Proved