Question

The elevation of a tower at a station A due north of it is "α" and at station B due west of A is "β". Prove that the height of tower is "ABsinαsinβ/√sin2α-sin2β"

Answer 1

Let O be the point where the tower stands.
Let A be the point due north of it and B the point due east of A.
From A the angle of elevation of the tower is α.
∴ OA / h = cot α ⇒ OA = h Cot α ----------→(1)
From B the angle of elevation of the tower is β.
∴Cot β  = OB / h ⇒ OB = h cot β -------------→(2)
Consider  ΔOAB
AB2 = OB2 - OA2
       = h2 Cot2 β - h2 cot2 α
       =h2 (Cos2 β / Sin2 β-  Cos2 α / Sin2 α)
       =h2 (Sin2 α Cos2 β -  Cos2 α Sin2 β) / Sin2 α Sin2 β.
       =h2 (Sin2 α (1-Sin2 β) -  (1-Sin2 α) Sin2 β) / Sin2 α Sin2 β.
AB2 =h2 (Sin2 α -  Sin2 β) / Sin2 α Sin2 β.
h2  =   AB2 Sin2 α Sin2 β./ (Sin2 α -  Sin2 β)
∴ h = AB Sin α Sin β./ √(Sin2 α -  Sin2 β).       

Answer 2

here is your correct answer