Question

The angle of elevation of the top of a tower from a point A due south of the tower is α and from B due east of tower is β. If AB = d, show that the height of the tower isd/âcot2α+ cot2β

Answer 1

Let D be the point where the tower stands.
Let A be the point due south of it and B the point due east of A.
From A the angle of elevation of the tower is α.
Distance between A and B is AB = d.
∴ AD / h = cot α ⇒ AD = h Cot α
AD2 = h2 Cot2 α ----------→(1)
From B the angle of elevation of the tower is β.
∴Cot β  = DB / h ⇒ DB = h cot β
DB2= h2 cot2 β -------------→(2)
Consider  Δ ADB
AB2 = AD2 + DB2
  d2  = h2 cot2 α + h2 Cot2 β
  d2  = h2 (cot2 α +  Cot2 β)
 h = d / √(cot2 α +  Cot2 β)
∴ height of the tower is = d / √(cot2 α +  Cot2 β).

An alternate method:

Let OP be the tower and let A and B be two points due south and east respectively of the 
tower such that ∠OAP = α and ∠OBP = β. Let OP = h.
In △OAP, we have 
tan α = h/OA 
OA = h cot α ........(i)
In △OBP, we have 
tan β = h/OB
OB = h cot β ........(ii)
Since OAB is a right angled triangle. Therefore,
AB2 = OA2+OB2
d2 = h2cot2 α + h2 cot2β
h = d/√cot2a + cot2β  [using (i) and (ii)].    

Hope This Helps :)