Question

A man observes a car from the top of a tower , which is moving towards the tower with a uniform speed . if the angle of depression of the car changes from 30 and 40 in 12 minutes., find the time taken by heads car now to reach the tower.

Answer 1

Hey mate

Here is ur answer


It must be 30 to 45 in the question. As the value for 40 is not given in the degree table.

Let A be the position of the man and AB be the tower.

Now consider the triangle ABD
tan 30 = AB/BD
1/√3 = AB/BD
BD/√3 = AB
BC+CD/√3 = AB......... (1)

Consider the triangle ABC
tan45 = AB/BC
1 = AB/BC
AB=BC.........(2)

Substitute the value of BC from eq(2) in eq(1)
AB+CD/√3 = AB
AB+CD = √3AB
√3AB-AB = CD
AB(√3-1) = CD....... (3)

Let v/m min be the speed of the car.
Since the car takes 12min to cover the distance CD, we have
CD= 12v......... (4) (distance=speed*time)

Substitute the value of CD from eq(4) in eq(3)
AB(√3-1)/12= v...... (5)
Time taken by the car to cover BC
= BC/v
Time = 12BC/BC(√3-1).....(FROM 5)
Time = 12AB/AB(√3-1)....(FROM 4)
Time = 12/√3-1)
= 16.4 minutes

Answer 2

Hey mate.....

here's ur answer

Hope it helps

Be brainly