Question

A straight highway leads to the foot of a Tower a man standing at the top of the tower of reserves a car at an angle of depression of 30 degree which is approaching the foot of the tower with a uniform speed 6 seconds letter the angle of depression of the car is found to be 60 degree find the time taken by the car to reach the foot of the tower from the point​

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 …..

Again, From right angled ∆ADC, tan 60° = AD/CD ⇒ √3 = h/CD ⇒ h = √3CD …..

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.

Answer 2

Answer:

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