A boy at top of a vertical observation tower observes a car moving at uniform speed coming directly towards the tower. If it takes 10 minutes for angle of depression to change from 30degree to 60 degree, how soon after this will the car reach the vertical tower
Answers
Hi dude✌.........
A man on top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 mins for the angle of depression to change from 30 degree to 45 degree........
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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