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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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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 6$x$ m. Let $t$ time required to reach the foot of the tower. So, BC = $x$$t$

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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