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

Answer: 18.65

In $\Delta QPA:\frac{h}{x+y}=tan{\left(30\right)}^o\Rightarrow \sqrt{3}h=x+y\dots \dots .(i)$
In $\Delta QPB:\frac{h}{y}=tan{\left(45\right)}^o\Rightarrow h=y\dots \dots .(ii)$
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