Question

The angle of elevation of top point P of vertical tower PQ of height h from point A is 45 and from point B is at a distance d from point A measure along the line AB which makes an angle 30 with AQ. Prove that d=h( Ö 3-1)

Answer 1

"∴ ∠ APM = 60° :

Also PN ⊥ AB, therefore AN = NM = 20 m

⇒ AP = 40 m

Let angles of elevation of top of the tower from A, N and B be α, θ and β respectively. ATQ, tan θ = 2

In ∆ PQN tan θ = PQ/PN

⇒ 2 = h/PN ⇒ PN = h/2 . . . . . . . . . . . . . . . . . . . (1)

Also in ∆APM, ∠APM = 60° (being equilateral ∆) and PN is altitude ∴ ∠APN = 30°(as in equilateral ∆

altitude bisects the vertical angle.

∴ In ∆APN tan ∠ APN = AN/PN

⇒ tan 30°= 20 / h/2 [Using eq. (1)]

⇒ h/2√3 = 20 ⇒ h = 40√3m.

In ∆APQ tan α = h/AP ⇒ tan α = 40√3/40 = √3

⇒ α = 60° Also in ∆ABQ tan β = h/PB but in rt ∆PNB

∴ PB = √1200 + 3600 = √4800 = 40 √3

∴ tan β = 40 √3/40 √3 ⇒ tan β = 1 ⇒ β = 45°

Thus h = 40 √3m ; ∠’s of elevation are 60°, 45°

"