A ladder rest 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. Prove that `x=ytan(alpha+beta)/2dot`

Options

A. `y= xtan""(alpha+beta)/2`
B. `x= ytan""(alpha+beta)/2`
C. `x= ytan""(alpha+beta)`
D. `y= xtan""(alpha+beta)`

Explanation

Let AB be the wall, P be the intial position and Q be the final position of the foot of the ladder on the ground
In the figure , PB =QC =1
In triangle BAP,PA = l cos `alpha` AB= li sin `alpha`
In triangle CAQ , QA = l cos `beta` , AC l sin `beta `
Now `y= BC= AB - AC =(sin alpha - sin beta )`
and `= x =PQ = AQ - AP = l( cos beta - cos alpha)`
`rArr y/x=(sin alpha- sin beta)/(cos beta -cos alpha)`
`(2sin ""((alpha-beta)/2) cos"" ((alpha+beta)/2))/(2sin ""((alpha+beta)/2) cos"" ((alpha-beta)/2))`
`rArr y/x=cot ""((alpha +beta)/(2))`
`rArr y/x=cot ""((alpha +beta)/(2))`
`rArr x= y tan ""((alpha+ beta)/2)`