A boy on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 30 to 60", how soon after this will the car reach the vertical tower.
Answers
Answer:
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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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