Question

A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards him. If it takes 10 minutes for the angle of depression to change from 60 degree to 45 degree, how soon after this, will the car reach the tower? Give your answer to the nearest second.​

Answer 1

13.69 minutes

Step-by-step explanation:

See the attached diagram.

From the right triangle Δ ABC, $\tan 45^{\circ} = \frac{AB}{AC} = \frac{h}{x + y}$

⇒ h = x + y ........... (1)

Now, from the right triangle Δ ABD, $\tan 60^{\circ} = \frac{AB}{AD} = \frac{h}{x}$

⇒ x = 0.58h (Approx.) ........... (2)

So, from equation (1) we get,

y = h - x = h - 0.58h = 0.42h

Now, the car travels 0.42h distance in 10 minutes i.e. 0.167 hours.

So, the speed of the car is, $v = \frac{y}{0.167} = \frac{0.42h}{0.167} = 2.53h$ .

Now, the time required by the car to reach the tower bottom is $\frac{x}{v} = \frac{0.58h}{2.53h} = 0.223$ hours = 13.69 minutes. (Answer)

Answer 2

Solution :-

in Right ∆ABC, we have,

→ tan 45° = AB / BC

→ 1 = AB / BC

→ AB = BC --------- Eqn.(1)

now, in Right ∆ABD , we have,

→ tan 30° = AB / BD

→ (1/√3) = AB / (AB + CD)

→ (1/√3) = AB/(AB + CD)

→ √3AB = AB + CD

→ √3AB - AB = CD

→ AB(√3 - 1) = CD ------------ Eqn.(2)

now, Let us assume that, speed of car is x m/min.

so,

→ Distance covered by car in 10min. = CD

→ Speed * Time = CD

→ 10x = CD

→ x = CD/10 ---------------- Eqn.(3)

therefore,

→ Time taken by car to reach tower = Distance / speed

→ Time = BC / x

putting value of Eqn.(3),

→ Time = 10BC / CD

putting value of Eqn.(2) now,

→ Time = 10BC/AB(√3 - 1)

Putting value of Eqn.(1) now,

→ Time = 10AB / AB(√3 - 1)

→ Time = 10/(√3 - 1)

→ Time = 10/(1.73 - 1)

→ Time = 10/(0.73)

→ Time ≈ 13.7 minutes (Ans.)

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