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. Show that cos cos
sin sin
a
b
α β
β α
−
=
−

Answer 1

In ∆ OQB, cos β = OB/BQ

⇒ OB = ℓ cos β . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)

Similarly in ∆ OPA, cos α = OA/PA

⇒ OA = ℓ cos α . . . . . . . . . . . . . . . . . . . . . . . . . (2)

Now a = OB – OA = ℓ (cos β – cos α) . . . . . . . . . . . . . . . . . . (3)

Also from ∆ OAP, OP = ℓ sin α

And in OQB; OQ = ℓ sin β

∴ b = OP – OQ = ℓ (sin α – sin β) . . . . . . . . . . . . . . . . . . . . . . (4)

Dividing eq. (3) by (4) we get

a/b = cos β – cos α/sin α – sin β

\frac{2sin(\frac{\alpha + \beta}{2})sin(\frac{\alpha - \beta}{2})}{2cos(\frac{\alpha +\beta}{2})-sin(\frac{\alpha-\beta}{2})}

⇒ a/b = tan (α + β/2)

Thus , a = b tan (α + β/2) is proved