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Question
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution 1
Let AB be the tower.
Initial position of the car is C, which changes to D after six seconds.
In ΔADB,
`("AB")/("DB") = tan 60º`
`("AB")/("DB") =sqrt3`
`"DB" = ("AB")/sqrt3`
In ΔABC,
`("AB")/("BC") = tan 30º`
`("AB")/("BD" + "DC") = 1/sqrt3`
`"AB"sqrt3 = "BD" + "DC"`
`"AB"sqrt3 = ("AB")/sqrt3 + "DC"`
`"DC" = "AB"sqrt3 - ("AB")/sqrt3 = "AB"(sqrt3 - 1/sqrt3)`
= `(2"AB")/sqrt3`
Time taken by the car to travel distance DC `("i.e" "2AB"/sqrt3)` = 6 second
Time taken by the car to travel distance DB `("i.e" ("AB")/sqrt3) = 6/((2"AB")/sqrt3)xx("AB")/sqrt3`
= `6/2`
= 3 seconds
Solution 2
Let PQ be the tower.
We have,
∠PBQ = 60° and ∠PAQ = 30°
Let PQ = h, AB = x and BQ = y
In ΔAPQ,
`tan 30° = (PQ)/(AQ)`
⇒ `1/ sqrt(3) = h/(x+y) `
⇒ `x+y = h sqrt(3)` ...(1)
Also, in ΔBPQ,
`tan 60° = (PQ)/(BQ)`
⇒ `sqrt(3) = h/y`
⇒ `h = y sqrt(3) ` ...(2)
Substituting `h = y sqrt(3)` in (i), we get
`x +y = sqrt(3) (ysqrt(3))`
⇒ x + y = 3y
⇒ 3y - y = x
⇒ 2y = x
⇒ `y = x/2`
`"As, speed of the car from "A to B = (AB) /6 = x/6 "units"/ "sec"`
So, the time taken to reach the foot of the tower i.e. Q from B `(BQ)/(speed)`
= `y/((x/6))`
= `((x/2))/((x/6))`
= `6/2`
= 3 sec
So, the time taken to reach the foot of the tower from the given point is 3 seconds.
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