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A ladder rests against a vertical wall at an inclination α 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 β with horizontal. State whether if LHS equals to RHS for \frac{p}{q}=\frac{\cot\beta-\cos\alpha}{\sin\alpha-\sin\beta}.

Answer:

Consider a vertical wall WB. Two positions AW and LD of a ladder as shown in figure such that LA = p, WD = q and LD = AW = z. Angle of inclination of ladder at two positions A and L are α and β respectively.

Let AB = y and DB = x.

In ΔABW, we have

sin α = (x+q)/z

and cos α = y/z

In ΔLBD, we have

Sin β = x/z

and cos β = (y+p)/z

\text { Taking RHS = } \frac{\cos \beta-\cos \alpha}{\sin \alpha-\sin \beta} \\ =\frac{\frac{y+p}{z}-\frac{y}{z}}{\frac{x+q}{z}-\frac{x}{z}}=\frac{\frac{y+p-y}{z}}{\frac{x+q-x}{z}} \\ =\frac{p}{z} \div \frac{q}{z}=\frac{p}{z} \times \frac{z}{q} \\ =\frac{p}{q}=\mathrm{LHS}

Hence, proved.

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