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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°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car

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°, 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.

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M manish


Let h be the height of the tower (DC) and the speed of the car be x\ ms^{-1}. Therefore, the distance (AB)covered by the car in 6 seconds is 6x m. Let t time required to reach the foot of the tower. So, BC = xt

According to question,
In triangle \Delta DAC,
\\\tan 30^0 = \frac{1}{\sqrt{3}}=\frac{h}{6x+xt}\\ x(6+t) = h\sqrt{3}..........................(i)

In triangle \Delta BCD,

\\\tan 60^0 = \sqrt{3} = \frac{h}{xt}\\ \therefore h = 3.xt...................(ii)

Put the value of h in equation (i) we get,
\\x(6+t) = (\sqrt{3}.\sqrt{3})xt\\ 6x +xt = 3xt\\ 6x = 2xt
t = 3

Hence, from point B car take 3 sec to reach the foot of the tower.

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