Heights and Distances Practice Test 2

Let AB = CD = h m be the heights of the towers. E is a point such that DE = 100m; ∠CED = 60° and ∠AEB = 30°
Now, BE = x m (say)
From right triangle CDE,
h = 100 tan 60° = 100√3 m
From right triangle ABE,
x = h cot 30° = 100√3×√3 = 300 m
Distance between the towers = DE + EB = 100 + 300 = 400 m
Height of the tower =100√3 m