A straight highway leads to the foot of a tower. A man standing on 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 form this point.
Let PQ be the tower.
We have,
$\angle \mathrm{PBQ}=60^{\circ}$ and $\angle \mathrm{PAQ}=30^{\circ}$
Let $\mathrm{PQ}=h, \mathrm{AB}=x$ and $\mathrm{BQ}=y$
In $\triangle \mathrm{APQ}$,
$\tan 30^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AQ}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+y}$
$\Rightarrow x+y=h \sqrt{3} \quad \ldots$ (i)
Also, in $\triangle \mathrm{BPQ}$,
$\tan 60^{\circ}=\frac{\mathrm{PQ}}{\mathrm{BQ}}$
$\Rightarrow \sqrt{3}=\frac{h}{y}$
$\Rightarrow h=y \sqrt{3} \quad \ldots$ (ii)
Substituting $h=y \sqrt{3}$ in (i), we get
$x+y=\sqrt{3}(y \sqrt{3})$
$\Rightarrow x+y=3 y$
$\Rightarrow 3 y-y=x$
$\Rightarrow 2 y=x$
$\Rightarrow y=\frac{x}{2}$
As, speed of the car from A to B $=\frac{\mathrm{AB}}{6}=\frac{x}{6}$ units / sec
So, the time taken to reach the foot of the tower i.e. $\mathrm{Q}$ from $\mathrm{B}=\frac{\mathrm{BQ}}{\text { speed }}$
$=\frac{y}{\left(\frac{x}{6}\right)}$
$=\frac{\left(\frac{x}{2}\right)}{\left(\frac{x}{6}\right)}$
$=\frac{6}{2}$
$=3 \mathrm{sec}$
So, the time taken to reach the foot of the tower from the given point is 3 seconds.
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