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Home » 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.
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.
CBSE | Class 10 | Maths | NCERT | Some Applications Of Trigonometry
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.

Solution:

Let the tower be AB.

The initial position and final position of the car be D and C respectively.

Since the man is at the top of the tower, as a result the angles of depression are measured from A.

The distance from the foot of the tower to the car = BC

Step 1: In right angle ΔABC,

AB/BC = tan 60°

AB/BC = √3

AB/√3= BC

AB = √3 BC

Step 2:

In right angle ΔABD,

AB/BD = tan 30°

AB/BD = 1/√3

AB = BD/√3

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

Since LHS are same, so RHS are also same, i.e. √3 BC = BD/√3

3 BC = BD

3 BC = BC + CD

2BC = CD

or BC = CD/2

BC is half the distance of CD in this case. As a result, the time taken is also half.

As a result, the time taken by the car to travel distance CD is 6 sec and the time taken by the car to travel BC is 6/2 sec. i.e. 3 sec.

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