A straight highway leads to the foot of a tower. A man standing on its top observes a car at an angle of depression of 30^o, which is approaching the foot of tower with a uniform speed. 6 seconds later, the angle of depression of the car becomes 60^o. Find the time taken by the car to reach the foot of tower from this point.

Given:

Angle of depression 1 (say ∠PAB) = 30^o

Angle of depression 2 (say ∠PAC) = 60^o

Time (say t) = 6 seconds

Let, Height of man and tower = AD

∵ the tower is vertical

∴ ∠ADB = 90°

Now,

PA∥BD

AB is transversal

∴∠ABD = ∠PAB = 30° [Alternate Angles]

Similarly,

PA∥BD

AC is transversal

∴∠ACD = ∠PAC = 60° [Alternate Angles]

In ∆ABD,

…(1)

Similarly,

In ∆ACD,

…(2)

From (1) and (2),

⇒ BD = √3×√3×CD

⇒ BD = 3CD

⇒ BC + CD = 3CD

⇒ BC = 3CD – CD

⇒ BC = 2CD

Now,

Time taken to cover BC = 6 seconds

Time taken to cover CD(1/2BC) =1/2× 6 seconds = 3 seconds

Hence, the time taken to reach the foot of tower from C = 3 seconds