A man of the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. if it takes 12 min. 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 this point ?
Answers
Answer: 16.4 minutes
Step-by-step explanation:
Let A be the position of the man and AB be the tower.
Now consider the triangle ,ABD
tan 30° = AB/BD
1/√3 =AB / BD
BD/ √3 = AB
BC + CD /√3 + AB....(1)
In Δ ABC
tan 45° = AB/BC
1 = AB/BC
AB =BC ..(2)
substitute the value of BC from equation (1) & (2)
Ab + CD/ √3 =Ab
Ab + CD =√3 AB
√3 AB -AB =CD
AB (√3 -1) = CD...(3)
The car takes 12 min to cover the distance CD
CD= 12v...(4)
Putting the value of CD in equation (4)
AB (√3-1) = 12v
AB (√3 -1)/12 = v...(5)
time taken by the car to cover the distance BC
Time = Distance / Speed
Time = BC/ AB (√3 -1 )
Time = 12 AB / AB (√3-1)
Time = 12 / √3-1
≈ 16.4 minutes
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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