Trigonometric Equations Question 32
Question: 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]
Options:
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 ) $
Show Answer
Answer:
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 ( \frac{\alpha -\beta }{2} ),\cos ( \frac{\alpha +\beta }{2} )}{2\sin ( \frac{\alpha +\beta }{2} ),\sin ( \frac{\alpha -\beta }{2} )} $
$ \Rightarrow $ $ \frac{y}{x}=\cot ( \frac{\alpha +\beta }{2} )\Rightarrow x=y\tan ( \frac{\alpha +\beta }{2} ) $ .
BETA
