(d) \(\frac{2\sqrt3+1}{6}\)
Let AB be the building of height h metres and BC the flagpole of height 50 ft.
Given, D is the position of the observer 300 ft away from the base of the building, i.e., AD = 300 ft.
Also ∠BDA = 30° and ∠CDA = θ
In rt. Δ ABD, tan 30° = \(\frac{AB}{AD}\)
⇒ \(\frac{h}{300}\) = \(\frac{1}{\sqrt3}\) ⇒ h = \(\frac{300}{\sqrt3}\) = 100√3 ft.
In rt. Δ ACD, tan θ = \(\frac{AC}{AD}\)
⇒ \(\frac{50+h}{300}\) = \(\frac{50+100\sqrt3}{300}\) = \(\frac{2\sqrt3+1}{6}\) m.