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

According to given information, we have the following figure

Now, from $\mathrm{△}ACD\text{ and }\mathrm{△}BCD\text{, }$ we have.

                      $\tan {30}^{\circ }=\frac{h}{x+y}$

and              $\tan {60}^{\circ }=\frac{h}{y}$

$\Rightarrow h=\frac{x+y}{\sqrt{3}}$                    (i)

and            $h=\sqrt{3}y$                      (ii)

From Eqs. (i) and (ii),

$\begin{array}{l}\frac{x+y}{\sqrt{3}}=\sqrt{3}y\\ \Rightarrow x+y=3y\\ \Rightarrow x-2y=0\Rightarrow y=\frac{X}{2}\end{array}$

$\therefore$Speed is uniform.

$\therefore$Distance y will be cover in 5 min.

$\therefore$Distance x covered in 10 min.

$\therefore$Distance $\frac{x}{2}$will be cover in 5 min.