A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. From the top of T2, if the angle of depression of the foot of T1 is twice the angle of elevation of the top of T2 from the top of T1, then find the width of the road between the feet of the towers T1 and T2. (Use tan2θ = 2tanθ / (1 - tan²θ))

Let the width of the road between the feet of the towers T1 and T2 be w.
In △ABC, tanθ = BC/AC ⇒ tanθ = 20/w …(i)
Now, in △BOD, tan2θ = BO/OD ⇒ tan2θ = 80/w …(ii)
Now, by using the given formula of tan2θ in the question, tan2θ = 2tanθ / (1 - tan²θ)
80/w = 2 × (20/w) / (1 - (20/w)²)
[1 - (20/w)²] × 80/w = 40/w
1 - (20/w)² = 1/2
(20/w)² = 1/2
20/w = 1/√2
w = 20√2 m
Hence, the distance between the two towers is equal to 20√2 m.