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A straight highway leads to the foot of a tower. A man standing on the top of the $75$ m high tower observes two cars at angles of depression of $30^\circ$ and $60^\circ$, which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (use $\sqrt{3} = 1.73$)
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Sol.
AB = Height of tower = $75$ m
P, Q are positions of cars
$\angle XBQ = \angle BQA = 30^\circ$
$\angle XBP = \angle BPA = 60^\circ$
In $\Delta APB$, $\tan 60^\circ = \frac{75}{AP} \Rightarrow AP = \frac{75}{\sqrt{3}} = 25\sqrt{3}$
In $\Delta AQB$, $\tan 30^\circ = \frac{75}{AQ} \Rightarrow AQ = 75 \sqrt{3}$
Distance between the cars = $PQ = AQ – AP$
$= 75\sqrt{3}-25\sqrt{3} = 50\sqrt{3}$
$= 50 \times 1.73 = 86.5$ m
AB = Height of tower = $75$ m
P, Q are positions of cars
$\angle XBQ = \angle BQA = 30^\circ$
$\angle XBP = \angle BPA = 60^\circ$
In $\Delta APB$, $\tan 60^\circ = \frac{75}{AP} \Rightarrow AP = \frac{75}{\sqrt{3}} = 25\sqrt{3}$
In $\Delta AQB$, $\tan 30^\circ = \frac{75}{AQ} \Rightarrow AQ = 75 \sqrt{3}$
Distance between the cars = $PQ = AQ – AP$
$= 75\sqrt{3}-25\sqrt{3} = 50\sqrt{3}$
$= 50 \times 1.73 = 86.5$ m