A man on the top

Let AB be the tower with height of tower AB = h

At C the angle of depression of car measures 30 and 12 minutes later it reaches D where angle of depression is 45.

Let CD = x, D = y

Here, AB = h, ∠ACB = ∠XAC = 30⁰

∠ADB = ∠XAD = 45⁰

In ∆ACB,

x + y = √3h ……(1)

In ∆ABD,

H = y ………(2)

From equation 1 and 2,

x + y = √3y

x = (√3 – 1)y

The distance covered by car in 12 minutes is CD = x

So, time taken to cover distance x is 12 minutes.

So, y = Time taken to cover distance DB

= 6 × 2.73

= 16.38 minutes

The time taken by car to reach the tower is 16.38 minutes.