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 degree to 45 degree how long will the car take to reach the observation Tower from the point on this.
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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