Angle between the hour and minute hands at 6:05

Think of it this way: five minutes after six, the minute hand is $\frac1{12}$ of the circle ahead from 12, while the hour hand has advanced $\frac1{12}$ of the way towards 7 from 6, or $\frac1{144}$ of the circle ahead. The initial angle between the two hands is $\frac12$ of the circle, so the solution is $$\frac12-\frac1{12}+\frac1{144}=\frac{61}{144}=152.5^\circ$$

Here is how much the hour hand travels per hour, minute, second:

• Per Hour: $$\text{H}=\frac{360^{\circ}}{12\space\text{hours}}=30^{\circ}\text{/}\space\text{hour}$$
• Per Minute: $$\text{M}=\frac{\text{H}^{\circ}}{60\space\text{minutes}}=\left(\frac{1}{2}\right)^{\circ}\text{/}\space\text{minute}$$
• Per Second: $$\text{S}=\frac{\text{M}^{\circ}}{60\space\text{seconds}}=\left(\frac{1}{120}\right)^{\circ}\text{/}\space\text{second}$$

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So, when it is $6:05$, we get:

$$6\cdot30^{\circ}+5\cdot\left(\frac{1}{2}\right)^{\circ}=182.5^{\circ}$$

But, for the 'minute hand' we got $30^{\circ}$ too much, so:

$$\text{angle}=182.5^{\circ}-30^{\circ}=152.5^{\circ}$$

Your approach is correct, but the hour is $6+\frac{5}{60}=\frac{73}{12}$, not $\frac{73}{2}$.

This yields an angle of $\frac{73}{12}\cdot30^\circ$ for the hour hand, and $6^\circ\cdot 5=30^\circ$ for the minute hand.

Thus the end result is $\frac{73}{12}\cdot30^\circ-30^\circ=152.5^\circ$