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 ΔOAP, we have

tan α = h / OA

OA = h cot α ……………. (i)

In ΔOBP, we have

tan β = h / OB

OB = h cot β ……………… (ii)

Since OAB is a right angled triangle.

Therefore,

AB ² = OA ² + OB ²

d ² = h ² cot ² α + h ² cot ² β

h = [Using (i) and (ii)].