A man is walking towards a vertical pillar in a straight path at a...

In $∆ QPA: \dfrac{h}{x + y} = tan {30}^{o} ⇒ \sqrt{3}h = x + y \ldots \ldots .(i)$
In $∆ QPB: \dfrac{h}{y} = tan {45}^{o} ⇒ h = y \ldots \ldots .(ii)$
From (i) and (ii), we get,

$\sqrt{3} y = x + y ⇒ y = \dfrac{x}{\sqrt{3} - 1}$
Given that the distance $x$ is covered in $5$\left(\sqrt{3} + 1\right)$minutes<br />Hence, the distance$\dfrac{x}{\sqrt{3} - 1}$is covered in$\dfrac{5\left(\sqrt{3} + 1\right)}{\left(\sqrt{3} - 1\right)} = 5\left(2 + \sqrt{3}\right) = 18 . 65 minutes