Question

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. The value of a/b is

(a) \(tan\, \left(\frac{\alpha\,+\,\beta}{\alpha\,-\,\beta}\right)\)

(b) \(tan\, \left(\frac{\alpha\,-\,\beta}{\alpha\,+\,\beta}\right)\)

(c) \(tan\, \left(\frac{\alpha\,-\,\beta}{2}\right)\)

(d) \(tan\, \left(\frac{\alpha\,+\,\beta}{2}\right)\)

Answer 1

(d) \(tan\, \left(\frac{\alpha\,+\,\beta}{2}\right)\) 

Let OY be the wall and OX the horizontal. 

Let AB be the initial position of the ladder and PQ the new position of the ladder after its foot is pulled away from the wall. Let the length of the ladder be x metres.

Given,

QB = a m, AP = b m,

AB = PQ = x m, 

∠ABO = α, ∠PQO = β

In rt. Δ AOB,

cos α = \(\frac{OB}{AB}\)

⇒ OB = AB cos α = x cos α

and sin α = \(\frac{OA}{AB}\)

⇒ OA = AB sin α = x sin α

In rt. Δ POQ,

cos β = \(\frac{OP}{PQ}\) 

⇒ OP = PQ cos β = x cos β

and sin β = \(\frac{OP}{PQ}\)

⇒ OP = PQ sin β = x cos β

a = OQ – OB 

= x cos β – x cos α 

= x (cos β – cos α)

b = OA – OP 

= x sin α – x sin β 

= x (sin α – sin β)

∴ \(\frac{a}{b}=\frac{cos\,\beta\,-\,cos\,\alpha}{sin\,\alpha\,-sin\,\beta}\)

= \(tan \left(\frac{\alpha\,+\,\beta}{2}\right)\)