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 × 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)/sqrt(3)`   ...(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(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1))`

t = `(12(sqrt(3) + 1))/(3 - 1)`

t = `(12(sqrt(3) + 1))/2`

t = `6(sqrt(3) + 1)`

t = 15.39

∴ Time = 16.39 minutes

Time = 16 minutes 23 seconds.