A man at top of a vertical observation tower observes a car moving at uniform speed coming directly towards the tower. If it takes 12 minutes for angle of depression to change from 30degree to 45 degree, how much 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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