Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β, then tan β is equal to.

Let AB = x. Then AP = 2AB = 2x
△ABP is a right-angled triangle with BP as hypotenuse.
By Pythagoras Theorem, BP² = AP² + AB²
BP² = (2x)² + x²
BP² = 5x²
⇒ BP = √5x
AC = x/2
tan α = x/(2x) = 1/4
Now, tan(α + β) = x/x = 1/2
tan α + tan β / 1 - tan α tan β = 1/2
⇒ 2(tan α + tan β) = 1 - tan α tan β
⇒ 2(1/4 + tan β) = 1 - (1/4)tan β
⇒ 1/2 + 2tan β = 1 - (1/4)tan β
⇒ 9/4 tan β = 1/2
⇒ tan β = 2/9