Question

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 $5\left(\sqrt{3} + 1\right)$ minutes from $A$ in the same direction, at a point $B$ , he observes that the angle of elevation of the top of the pillar is $45^{o}.$ Then the time taken (in minutes) by him, to reach from $B$ to the pillar, is (take $\sqrt{3}=1.73$ )

Answer 1

In $\Delta QPA:\frac{h}{x + y}=tan \left(30\right)^{o}\Rightarrow \sqrt{3}h=x+y\ldots ….(i)$
In $\Delta QPB:\frac{h}{y}=tan \left(45\right)^{o}\Rightarrow h=y\ldots \ldots .(ii)$
From (i) and (ii), we get,
$\sqrt{3} y = x + y \Rightarrow y = \frac{x}{\sqrt{3} - 1}$
Given that the distance $x$ is covered in $5 \left(\sqrt{3} + 1\right)$ minutes
Hence, the distance $\frac{x}{\sqrt{3} - 1}$ is covered in $\frac{5 \left(\sqrt{3} + 1\right)}{\left(\sqrt{3} - 1\right)}=5\left(2 + \sqrt{3}\right)=18.65$ minutes