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Question
A man on the top of a tower observes that a car is moving directly at a uniform speed towards it. If it takes 720 seconds for the angle of depression to change from 30° to 45°, how soon will the car reach the observation tower?
Solution
Let AB be the tower .
Initial position of car is C , which changes to D after 720 seconds.
In ΔADB
`"AB"/"DB" = tan45^circ`
`"AB"/"DB" = 1`
DB = AB
In ΔABC
`"AB"/"BC" = tan 30^circ`
`"AB"/"BD + DC" = 1/sqrt(3)`
`"AB"sqrt(3) = "BD + DC"`
`"AB"sqrt(3) = "AB + DC"`
`"DC" = "AB"sqrt(3) - "AB" = "AB"(sqrt(3) - 1)`
Time taken by car to travel DC distance (i.e `"AB"(sqrt(3) - 1`)) = 720 seconds
Time taken by car to travel DB distance (i.e. AB)
= `720/("AB"(sqrt(3) - 1)) xx "AB" = 720/((sqrt(3) - 1)) xx (sqrt(3) + 1)/(sqrt(3) + 1)`
= `(720(sqrt(3) + 1))/2 = 360(sqrt(3) + 1) = 360 xx 2.732 = 983.52`
Thus , the required time taken is 983.52 seconds = 984 seconds = 16 mins 24 secs.
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