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A ladder rests against a wall with an angle $α$ $alpha$ to the horizontal. Its foot is being pulled away from the wall through a distance $a$ $a$ so that it slides $a$ $a$ distance $b$ $b$ down the wall making an angle $β$ $beta$ with the horizontal. Show that $a = b \cdot tan (\frac{α + β}{2})$ $a=b*tan((alpha+beta)/2)$

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Answer 1 · 2018-06-05T02:12:07

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Answer 2 · 2018-06-05T02:48:02

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Let AB and CD be 1st and 2nd positions of the ladder. So $A B = C D = x (s a y)$ $AB=CD =x (say)$ .
Again as per problem $B C = a and A D = b$ $BC=a and AD=b$

Now

$a = B C = E C - E B = (\frac{E C}{C D} - \frac{E B}{C D}) \cdot C D$ $a=BC=EC-EB=((EC)/(CD)-(EB)/(CD))*CD$

$\Rightarrow a = (\frac{E C}{C D} - \frac{E B}{A B}) \cdot C D$ $=>a=((EC)/(CD)-(EB)/(AB))*CD$

$=>a=(cosbeta-cosalpha)x...[1]$

Similarly

$b=AD=AE-ED=((AE)/(AB)-(ED)/(AB))*AB$

$=>b=((AE)/(AB)-(ED)/(CD))*AB$

$=>b=(sinalpha-sinbeta)x...[2]$

Dividing [1] by [2] we get

$a/b=(cosbeta-cosalpha)/(sinalpha-sinbeta)$

$=>a/b=(2sin((alpha+beta)/2)sin((alpha-beta)/2))/(2cos((alpha+beta)/2)sin((alpha-beta)/2))$

Hence

$a=btan((alpha+beta)/2)$

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