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A ladder rests against a wall at an angle $α$ 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 $β$ with the horizontal, then

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a = btan \frac{α + β}{2}

a = bcot \frac{α + β}{2}

atan \frac{α - β}{2}

None

answer is A.

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Detailed Solution

It is given that a ladder rests against a wall at an angle

α

to the horizontal.
According to the given data the figure has been drawn below.
Question Image

Here l is the length of the ladder i.e, BQ=AP=l.
In

∆ OQB,

cos θ = b a s e H y p o t e n u s e

cosβ = \frac{OB}{BQ}

⇒

OB = lcosβ . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)

Similarly in

∆ OPA,

cos θ = b a s e H y p o t e n u s e

cosα = \frac{OA}{PA}

\Rightarrow OA = lcosα . . . . . . . . . . . . . . . . . . . . . . . . . (2)

Now,

a = OB–OA

= l (cosβ–cosα) . . . . . . . . . . . . . . . . . . (3)

Also from

∆ OAP,

sin θ = P e r p e n d i c u l a r H y p o t e n u s e

sinα = \frac{OP}{PA}

\Rightarrow OP = lsinα

And in

△ OQB,

sin θ = P e r p e n d i c u l a r H y p o t e n u s e

sinβ = \frac{OQ}{PA}

\Rightarrow OQ = lsinβ

∴ b = OP–OQ

= l (sinα–sinβ) . . . . . . . . . . . . . . . . . . . . . . (4)

Dividing eq.

(3)

(4)

we get

\Rightarrow

\frac{a}{b} = \frac{cosβ–cosα}{sinα–sinβ}

\Rightarrow \frac{a}{b} = \frac{2 \sin \frac{α + β}{2} . \sin \frac{α - β}{2}}{2 \cos \frac{α + β}{2} . \sin \frac{α - β}{2}}

∵ cos A − cos B = 2 sin A + B 2 sin B − A 2 ∵ sin A + sin B = 2 sin A + B 2 cos A − B 2

\Rightarrow \frac{a}{b} = \frac{\sin \frac{α + β}{2}}{\cos \frac{α + β}{2}}

⇒

\frac{a}{b} = \tan \frac{(α + β)}{2}

∵ sin θ tan θ = tan θ

\Rightarrow a = b

\tan \frac{(α + β)}{2}

So the correct option is 1.

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