Question

A flagpole stands on a building and an observer on a level ground is 300 ft from the base of the building. The angle of elevation of the bottom of the flagpole is 30° and the height of the flagpole is 50 ft. If θ is the angle of elevation of the top of the flagpole, then tan q is equal to

(a) \(\frac{\sqrt3}{2}\)

(b) 1

(c) \(\frac{4\sqrt3+1}{6}\)

(d) \(\frac{2\sqrt3+1}{6}\)

Answer 1

(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.