(b) \(1/\sqrt6\)
Let PD be the pole standing at the corner D of the square plot ABCD.
Let PD = h m (say)
Given,
∠PCD = 30°, ∠DAP = 30° and ∠PBD = θ.
From rt. Δ PAD or Δ PCD PD = AD tan 30° or PD = CD tan 30°
Let each side of the square ABCD be a metres.
(Then, PD = a tan 30° = \(\frac{a}{\sqrt3}\))
BD being the diagonal of square ABCD,
BD = \(a\sqrt2\)
∴ In rt. Δ PBD,
tan θ = \(\frac{PD}{BD}\) = \(\frac{\frac{a}{\sqrt3}}{\frac{a}{\sqrt2}}\) = \(\frac{1}{\sqrt6}\).