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# A vertical tower PQ stands at a point P. Points A and B are located to the South and East of P respectively. M is the mid point of  AB PAM is an equilateral triangle; and N is the foot of the perpendicular from P on AB. Let AN = 20 meters and the angle of elevation of the top of the tower at N is tan-1 (2). Determine the height of the tower and the angles of elevation of the top of the tower at A and B.

Jitender Pal
7 years ago
Hello Student,
Let PQ = h
As A and B are located to the south and east of P respectively,
∴ ∠APB = 90°. M is mid pt of AB. PAM is an equilateral ∆
∴ ∠ 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 = $\sqrt{PN^2+NB^2}$ = $\sqrt{(20\sqrt{3})^2+(60)^2}$
∴ 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°

Thanks
Jitender Pal