Answer
The distance is $d=\sqrt{{{m}^{2}}+{{n}^{2}}-mh}$
Work Step by Step
In a clock there are $30$ divisions from $9$ to $12$ and the angle is ${{90}^{\circ }}$, which means $1$ division of the clock is equal to $3{}^\circ $.
So, at $10$ ‘o’ clock the angle will be $60{}^\circ $.
We can use the below figure to represent the given situation. According to the cosine laws,
$\begin{align}
& {{d}^{2}}={{m}^{2}}+{{h}^{2}}-2mh\cos 60{}^\circ \\
& ={{m}^{2}}+{{h}^{2}}-2mh\left( \frac{1}{2} \right) \\
& ={{m}^{2}}+{{h}^{2}}-mh \\
& d=\sqrt{{{m}^{2}}+{{h}^{2}}-mh}
\end{align}$
Here $d$ is the distance between the tip of the minute and the hour sticks of a clock.
