Find the angle between hour hand and minute hand at 1:59? I have got the formula to find angle between hour hand and minute hand from   
http://en.wikipedia.org/wiki/Clock_angle_problem
The angle between the hands can be found using the formula:
$$\begin{align} \Delta\theta &= \left|\theta_{\text{hr}} - \theta_{\text{min.}}\right| \\ &= \left|\frac{1}{2}(60H + M) - 6M\right|\\ &= \left|\frac{1}{2}(60H - 11M)\right| \end{align}$$
where$    \scriptstyle H$ is the hour
  $  \scriptstyle M$ is the minute. 
According to this formula at 1:59, H=1, M=59  
$ \Delta\theta=|\dfrac{60\times 1-11\times 59}{2}|=294.5^\circ$ which is not true. What is the problem, please help.

I couldn't make myself understand the formula you point to. At 1:59 there are $\frac{360}{60}^o=6^o$ from the minute hand to noon. There are $\frac{360}{12}^o=30^o$ between neighboring hours, so there are $2\cdot30^o=60^o$ from noon to 2pm, total so far $66^o$. 
But the hour hand is not exactly at 2pm. It takes 1 hour for the hour hand to travel from 1pm to 2pm, i.e. 1 hour to travel $30^o$. So the hour hand in one minute will travel $\frac{30}{60}^o=\frac1{2}^o$. That is, the answer would be $66^o - \frac1{2}^o = \frac{132-1}{2}^o= \frac{131}{2}^o$. 
Since $\frac{131}{2}^o= 65.5^o$ this seems to agree with the answer from wikipedia, and from @Andrey. Indeed $65.5^o + 294.5^o = 360^o$ which confirms that the angle is the same, except when measured clockwise it is $294.5^o$, and when measured counterclockwise it is $65.5^o$. If the smaller angle is needed then the answer would be $65.5^o$. 
