Question

$\bf{Try\: this\: one...}$
A ladder rests against a vertical wall at an inclination $\alpha$ to the horizontal.Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle $\beta$ to the horizontal.Show that
$\frac{p}{q} = \frac{cos \beta - cos \alpha}{sin \alpha - sin \beta}$
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Answer 1

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Answer 2

$\large{\underline{\underline{\mathfrak{\sf{\red{Given-}}}}}}$

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A ladder rests against a vertical wall at an inclination $\alpha$ to the horizontal.Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle $\beta$ to the horizontal.

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$\large{\underline{\underline{\mathfrak{\sf{\red{To\:show-}}}}}}$

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• $\sf{\dfrac{p}{q} = \dfrac{cos \beta - cos \alpha}{sin \alpha - sin \beta}}$

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$\large{\underline{\underline{\mathfrak{\sf{\red{Proof-}}}}}}$

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Let the height of ladder be h.

$\implies$ XY = WZ = h

In ∆OXY,

$\implies$ $\sf{Sin\alpha=\dfrac{P}{H}}$

$\implies$ $\sf{Sin\alpha=\dfrac{OY}{XY}}$

$\implies$ $\sf{Sin\alpha=\dfrac{OY}{h}}$

By cross multiplying,

$\implies$ $\large{\boxed{\sf{OY=hSin\alpha}}}$

Now,

$\implies$ $\sf{Cos\alpha=\dfrac{B}{H}}$

$\implies$ $\sf{Cos\alpha=\dfrac{OX}{XY}}$

$\implies$ $\sf{Cos\alpha=\dfrac{OX}{h}}$

By cross multiplying,

$\implies$ $\large{\boxed{\sf{OX=hCos\alpha}}}$

In ∆OWZ,

$\implies$ $\sf{Sin\beta=\dfrac{P}{H}}$

$\implies$ $\sf{Sin\beta=\dfrac{OZ}{WZ}}$

$\implies$ $\sf{Sin\beta=\dfrac{OZ}{h}}$

By cross multiplying,

$\implies$ $\large{\boxed{\sf{OZ=hSin\beta}}}$

Now,

$\implies$ $\sf{Cos\beta=\dfrac{B}{H}}$

$\implies$ $\sf{Cos\beta=\dfrac{OW}{WZ}}$

$\implies$ $\sf{Cos\beta=\dfrac{OW}{h}}$

By cross multiplying,

$\implies$ $\large{\boxed{\sf{OW=hCos\beta}}}$

$\rule{200}3$

From figure,

• p = OW - OX

• q = OY - OZ

Taking its ratios :

$\sf{\dfrac{p}{q} = \dfrac{OW-OX}{OY-OZ}}$

$\implies$ $\green{\boxed{\sf{\dfrac{p}{q} = \dfrac{cos \beta - cos \alpha}{sin \alpha - sin \beta}}}}$

Hence proved!