A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30^o. After walking for 10 minutes from A in the same direction, at point B, he observes that the angle of elevation of the top of the pillar is 60^o. Then find the time taken (in minutes) by him, from B to reach the pillar.

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Detailed Solution

Given that, a man is walking towards a pillar.

At point A, the angle of elevation is 30^o.
Question Image

In $△ BCD,$
$\tan 60^{\circ} = \frac{h}{y}$
$\Rightarrow \sqrt{3} = \frac{h}{y}$

$\Rightarrow h = \sqrt{3} y \dots (i)$

In $△ ACD,$

$\begin{array}{l} \tan 30^{°} = \frac{h}{x + y} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{\sqrt{3} y}{x + y} \dots [By using (i)] \end{array}$

$\Rightarrow x + y = 3 y$
$\Rightarrow x = 2 y$
$\Rightarrow y = \frac{x}{2}$
A man walks x distance in 10 minutes.

The speed is uniform and not changing.
So, the time taken by a man to walk a distance will be 5 minutes.

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