question_answer

A ladder rests against a wall making an angle \[\alpha \]with the horizontal. The foot of the ladder is pulled away from the wall through a distance x, so that it slides a distance y down the wall making an angle\[\beta \]with the horizontal. The correct relation is   [IIT 1985]

A) \[x=y\tan \frac{\alpha +\beta }{2}\]

B) \[y=x\tan \frac{\alpha +\beta }{2}\]

C) \[x=y\tan (\alpha +\beta )\]

D) \[y=x\tan (\alpha +\beta )\]

Correct Answer: A

Solution :

\[PB=QC=l\] (Length of ladder) \[\Rightarrow \] \[PA=l\cos \alpha ,\,QA=l\,\cos \beta \] \[\Rightarrow \] \[AC=l\sin \beta ,\,AB=l\sin \alpha \] \[\Rightarrow \] \[CB=AB-AC=l\,(\sin \alpha -\sin \beta )\] \[\Rightarrow \] \[y=l(\sin \alpha -\sin \beta )\] and \[QP=x=AQ-AP=l\], \[(\cos \beta -\cos \alpha )\]     \[\Rightarrow \] \[\frac{CB}{QP}=\frac{\sin \alpha -\sin \beta }{\cos \beta -\cos \alpha }=\frac{y}{x}=\frac{2\sin \left( \frac{\alpha -\beta }{2} \right)\,\cos \left( \frac{\alpha +\beta }{2} \right)}{2\sin \left( \frac{\alpha +\beta }{2} \right)\,\sin \left( \frac{\alpha -\beta }{2} \right)}\] \[\Rightarrow \] \[\frac{y}{x}=\cot \left( \frac{\alpha +\beta }{2} \right)\Rightarrow x=y\tan \left( \frac{\alpha +\beta }{2} \right)\].