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.
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.