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 $30^{o}$ , 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 $60^{o}$ . Find the time taken by the car to reach the foot of the tower from this point.

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Solution

Let $D C = x, B C = y, A C = d$
$tan 30 = \frac{C O}{A C} = \frac{x}{d} = \sqrt{3}$ ....(1)
$tan 60 = \sqrt{3} = \frac{D C}{B C} = \frac{x}{y}$ ...(2)
from (1) and (2)
$d = \sqrt{3} x = \sqrt{3} (\sqrt{3} y) = 3 y$
$∴ s p e e d = \frac{d i s t a n c e}{t i m e}$
from A to B
speed $= \frac{d - y}{6}$
required
time $= \frac{distance (from B to c)}{speed}$
$= \frac{y}{\frac{d - y}{6}}$
$= \frac{6 y}{d - y}$ $(∵ d - 3 y)$
$= \frac{6 y}{3 y - y}$
$= \frac{6 y}{2 y}$
$= 3$ sec

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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 $30^{\circ}$ , 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.

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as 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 DC=x,BC=y,AC=dtan30=COAC=xd=√3 ....(1)tan60=√3=DCBC=xy ...(2)from (1) and (2)d=√3x=√3(√3y)=3y∴speed=distancetimefrom A to Bspeed =d−y6requiredtime =distance (from B to c)speed=yd−y6=6yd−y (∵d−3y)=6y3y−y=6y2y=3 sec