a Man observes a car from the top of a tower which is moving towards the tower with a uniform speed if the angle of depression of the car changes from 30 to 40 Degrees in 12 minutes find the time taken by the car now to reach the tower
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Let AB be the vertical tower. Suppose D and C be the positions of the car when the angle of depression from the top of the tower is 30° and 45° respectively.
Consider the uniform speed of car be vm/min.
Time taken for the angle of depression to change from 30° to 45° = 12 min
∠EAD = ∠ADB = 30° (Alternate angles)
∠EAC = ∠ACB = 45° (Alternate angles)
Suppose AB = h m and BC = x m.
CD = Distance covered by car in
In ΔABC,
In ΔADB,
Time taken by car to reach the tower from
∴ Time taken by car to reach the tower is 983 sec.
Consider the uniform speed of car be vm/min.
Time taken for the angle of depression to change from 30° to 45° = 12 min
∠EAD = ∠ADB = 30° (Alternate angles)
∠EAC = ∠ACB = 45° (Alternate angles)
Suppose AB = h m and BC = x m.
CD = Distance covered by car in
In ΔABC,
In ΔADB,
Time taken by car to reach the tower from
∴ Time taken by car to reach the tower is 983 sec.
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