a man on the top of a vertical tower observes a car moving towards the tower .if it takes 12 min for the angle of depression to change from 30 degree to 45 degree,how soon after ts car will reach the 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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