Question

A straight highway leads to the foot of the tower .A man standing at the top of the tower observes the car at an angle of depression of 30degree ,which is approaching the foot of the tower with the uniform speed .six seconds later ,the angle of depression of the car found is to be 60degree.find the time taken by the car to reach the foot of the tower from this point.
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Answer 1

Let BCD is a highway.

A tower is standing at point D of height h. From the top of tower of point A the angle of depression is 30°.

After 6 sec when car reaches at point C then angle of depression becomes 60°.

Hence distance covered in 6 sec = BC

From right angled ∆ADB,

tan 30° = AD/BD

⇒ 1/√3 = h/BD

⇒ BD = h√3 …..(i)

Again, From right angled ∆ADC,

tan 60° = AD/CD

⇒ √3 = h/CD

⇒ h = √3CD …..(ii)

Put the value of h in equation (i),

BD = √3CD × √3 = 3CD

BC + CD = 3CD

2CD = BC

CD = 1/2 BC

Since car is moving with uniform speed and distance CD is half of BC.

Hence, time taken to cover distance,

CD = 1/2 × time taken to cover distance

BC = 1/2 × 6

= 3 sec.

Hence, required time = 3 SEC's

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