Question

SOLVE__IT__

A QUESTION FROM HEIGHT AND DISTANCE ( CLASS- X)​

Answer 1

Question :

A ladder rests against a wall at an angle aplha to the horizontal.Its foot is pulled away from the wall through a distance a so that it slides a distance b down the wall making an angle beta with the horizontal .Show that

$\frac{a}{b} = \frac{cos \alpha - cos \beta }{sin \beta - sin \alpha }$

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Answer :

Let PQ be the leader such that it's top Q is on the wall OQ and bottom P is on the ground .The ladder is pulled away from the wall through a distance a so, that it's top Q slides and takes position N

Clearly PQ=NM

In ∆s POQ and PNMwe have

$sin \alpha = \frac{OQ \: }{PQ \: } cos \alpha = \frac{op \: }{PQ \: } sn \beta = \frac{On}{NM \: } cos \beta = \frac{om \: }{MN \: }$

$= > sin \alpha = \frac{b + y}{PQ \: } cos \alpha = \frac{x}{PQ \: } sin \beta = \frac{y}{mn} \: cos \beta = \frac{a +x \: }{PQ \: }$

$= > sin \alpha - sin \beta = \frac{b + y}{PQ} - \frac{y}{PQ} \: and \: cos \beta - cos \alpha = \frac{a + x}{PQ \: } - \frac{x}{PQ \: }$

$= > sin \alpha - sin \beta = \frac{b}{PQ \: } and \: cos \beta - cos \alpha = \frac{a}{?PQ \: }$

$= > \frac{sin \alpha - sin \beta }{cos \beta - cos \alpha } = \frac{b}{a}$

$= > \boxed{ \red{ \frac{a}{b} = \: \frac{cos \alpha - cos \beta }{sin \beta - sin \alpha } }}$

Proved ✓