# A ladder rests against a wall at angle α to the horizontal. Its foot is pulled away from the wall through a distance a, so that it sides a distance b down the wall making an angle β with the horizontal. Show that a = b tan 1/2 (α + β)

8 years ago
Hello Student,
Let the length of the ladder, then
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

Thanks