Question

A man on the top of vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower? (Give your answer correct to nearest seconds).

Answer 1

Here, ∠ACB = 30° and ∠ADB = 45°.
Let C denote the initial position of the car and D be its position after 12 minutes.
Let the speed of the car be x meter/minute, then
CD = 12x meters     .....( ∵ Distance = speed x Time)
Let the car take t minutes to reach the tower from D.
Then, DB = tx meters

Now in the right-angled triangles ACB,
tan 30° = `"AB"/"BD"`

⇒ `1/sqrt3 = "AB"/("BC" + "CD")`

⇒ `1/sqrt3 = "AB"/(12x + tx)`

⇒ `"AB" = (12x + tx)/sqrt3`         ....(1)

Also, in the right-angled triangle ADB,

tan 45° = `"AB"/"DB"`

⇒ 1 = `"AB"/"DB"`

⇒ AB = DB = tx    ......(2)

From (1) and (2), we have

t = `12/(sqrt3 - 1) = 12(sqrt3 + 1)/2`

t = `6(sqrt3 + 1)`

t = 15.39

∴ Time = 16.39 minutes
Time = 16 minutes 23 seconds.