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A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed.

  

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A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

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Let AB be the tower.

D is the initial and C is the final position of the car respectively.

BC is the distance from the foot of the tower to the car.

Step 1: In right ΔABC,

tan 60° = AB/BC

√3 = AB/BC

BC = AB/√3

AB = √3 BC

Step 2:

In right ΔABD,

tan 30° = AB/BD

1/√3 = AB/BD

AB = BD/√3

Step 3: Form step 1 and Step 2, we have

√3 BC = BD/√3 (Since LHS are same, so RHS are also same)

3 BC = BD

3 BC = BC + CD

2BC = CD

or BC = CD/2

Time taken by car to travel distance CD = 6 sec.

Time taken by car to travel BC = 6/2 = 3 sec.

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