## Precalculus (6th Edition) Blitzer

The distance is $d=\sqrt{{{m}^{2}}+{{n}^{2}}-mh}$
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.