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

ΔQPA: \dfrac{h}{x + y} = tan{30}^{°}

$\to \sqrt{3}h = x + y$ ......\left(i\right)$<br /><br />$ΔQPB: \dfrac{h}{y} = tan{60}^{°}$</p><p>$\to h = \sqrt{3}y$......\left(ii\right)$

From equation \left(i\right) \text{$and$} \left(ii\right),$we get</p><p>$3y = x + y$</p><p>$\to y = \dfrac{x}{2}$<br /><br />$∵$Speed is uniform<br /><br />Therefore, distance$x$walked by him in$10min .

and distance \dfrac{x}{2} walked by him in 5min .