Answer is (C) 3l/2
Let PQ is pillar of height h.
Let point B is at x distance from foot of pillar Q.
According to question,
∠PAQ = 30°, ∠PBQ = 60° and AB = l
From right angled ΔPQB,
tan 60° = PQ/BQ
⇒ √3 = h/x
⇒ h = √3 x …..(i)
From right angled ΔPQA,
tan 30° = PQ/AQ
⇒ 1/√3 = h/(l + x)
⇒ l + x = √3h
⇒ l + x = √3(√3x) [From equation (i) h = √3x]
⇒ l + x = 3x
⇒ 2x = l
⇒ x = l/2
Hence, distance between foot of pole and point A.
= AQ = AB + BQ
= l + x
= l + l/2
= 3l/2