The straight highway leads to the foot of the tower. A man standing $t$ the foot of the tower observes a car at an angle of depression of $30^{2}$ , which is approaching the foot of the tower with a uniform speed. 15 seconds later,the angle of the depression found to be $45^{2}$ . Find the further time taken by the car to reach the foot of the tower.

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Solution

Let speed be x km/min.
$tan 30^{^{^{\circ}}} = \frac{h}{h + y} = \frac{1}{\sqrt{3}} . . . (i)$
$\begin{matrix} s p e e d = x = \frac{y}{15} 15 = \frac{y}{x} \end{matrix}$
$s p e e d = x = \frac{y}{15} = \frac{h}{t} . . . (i i)$
$\begin{matrix} f r o m (i) \sqrt{3} h = h + y y = (\sqrt{3} - 1) h \frac{(\sqrt{3} - 1) h}{15} = \frac{h}{t} t = \frac{15}{(\sqrt{3} - 1)} sec \end{matrix}$

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Question 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 $30^{\circ}$ , 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.

Q. A straight highway leads to the foot of the 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 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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Let speed be x km/min.tan30∘=hh+y=1√3...(i)speed=x=y1515=yxspeed=x=y15=ht...(ii)from(i)√3h=h+yy=(√3−1)h(√3−1)h15=htt=15(√3−1)sec