A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^{\circ}$ , which is approaching the foot of the tower with a uniform speed. 12 seconds later, the angle of depression of the car is found to be $60^{\circ}$ . Find the time in seconds taken by the car to reach the foot of the tower from this point.

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Solution

The correct option is A
6

Let DB = x

In $Δ$ BDC,

BC = $x \sqrt{3}$

In $Δ$ ABC,

AB = $B C \sqrt{3}$ = 3x

Hence, AD = AB - DB = 2x

If a car takes 12s to travel 2x, then it will take 6s to travel x.

Hence car reaches the tower in another 6 seconds.

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Question 15
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^{\circ}$ , 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 $60^{\circ}$ . Find the time taken by the car to reach the foot of the tower from this point.

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The correct option is A 6 Let DB = x In ΔBDC, BC = x√3 In ΔABC, AB = BC√3 = 3x Hence, AD = AB - DB = 2x If a car takes 12s to travel 2x, then it will take 6s to travel x. Hence car reaches the tower in another 6 seconds.

The correct option is A
6

Let DB = x

In $Δ$ BDC,

BC = $x \sqrt{3}$

In $Δ$ ABC,

AB = $B C \sqrt{3}$ = 3x

Hence, AD = AB - DB = 2x

If a car takes 12s to travel 2x, then it will take 6s to travel x.

Hence car reaches the tower in another 6 seconds.