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Question
A man on the top of a tower observes a truck at an angle of depression ∝ where `∝ = 1/sqrt(5)` and sees that it is moving towards the base of the tower. Ten minutes later, the angle of depression of the truck is found to `β = sqrt(5)`. Assuming that the truck moves at a uniform speed, determine how much more ti me it will take to each the base of the tower?
Solution
In the figure , AB is the tower . A is the position of the man. C and D are the two positions of the truck.
Let the speed of the truck be x m/sec
Distance CD = Speed × time = 600x
In right triangle ABC,
`tan α = "h"/"BC"`
It is a given that `tan α = 1/sqrt(5)`
`"BC" = "h"sqrt(5)` .. (1)
In right triangle ABD,
`tan β = "h"/"BD"`
It is given that `tan β = sqrt(5)`
`"h" = sqrt(5)"BD"`
Now , CD = BC - BD
600x = 5BD - BD
BD = 150x
Time taken = `(150"x")/"x"` = 150 seconds
Thus , the time taken by the truck to reach the tower is 150 sec = 2 min 30 sec.
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