a man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it if it takes 12 minutes for the angle of depression to change from 30° to 45° how soon after this will the car reach the observation tower
Answers
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 12min. = CD
→ Speed * Time = CD
→ 12x = CD
→ x = CD/12 ---------------- Eqn.(3)
therefore,
→ Time taken by car to reach tower = Distance / speed
→ Time = BC / x
putting value of Eqn.(3),
→ Time = 12BC / CD
putting value of Eqn.(2) now,
→ Time = 12BC/AB(√3 - 1)
Putting value of Eqn.(1) now,
→ Time = 12AB / AB(√3 - 1)
→ Time = 12/(√3 - 1)
→ Time = 12/(1.73 - 1)
→ Time = 12/(0.73)
→ Time ≈ 16.4 minutes (Ans.)
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