Question

A man on the top of a tower standing on the seashore find the a boat coming towards him takes $10$ minutes for the angle of depression to change for ${30}^0$ to ${60}^0$ Find the time(in mimutes) taken by the boat to reach the shore from this position

Answer 1

Let $AB$ be the tower and $C$ and $D$ be the two position of the boat
Let height of tower $AB=h$, distance covered by boat $CD=x$ and remaining distance $AD=y$.
$\frac{h}{y}=\tan {60}^0=\sqrt{3}\Rightarrow y=\frac{h}{\sqrt{3}}$
$\frac{h}{x+y}=\tan {30}^0=\frac{1}{\sqrt{3}}\Rightarrow x+y=\sqrt{3}h$
$\therefore x=(x+y)-y=(\sqrt{3}h-\frac{h}{\sqrt{3}})=\frac{2h}{\sqrt{3}}$
Given that distance of $\frac{2h}{\sqrt{3}}$ is covered in $10min$
Remaining distance $y=\frac{h}{\sqrt{3}}$ will be covered in $5min$, as $y$ is half of $x$.