Angle between the hour and minute hands at 6:05 
What is the angle between the hour and minute hands of a clock at 6:05?

I have tried this
Hour Hand:
12 hour = 360°
1 hr = 30°
Total Hour above the Clock is $\frac{73}2$ hours
In Minute Hand:
1 Hour = 360°
1 minutes = 6°
Total Minutes covered by $6\times 5= 30$
$\frac{73}{2} \cdot30-30=345^\circ$ Is it Correct? 
 A: 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 $\frac{73}{12}\cdot30^\circ-30^\circ=152.5^\circ$
A: The hours hand revolves $360°/(12\cdot60)=0.5°$ per minute and the minutes hand $360°/60=6°$ per minute, so that the angle increases by $5.5°$ per minute.
Hence working modulo $360°$, $$(6\cdot60+5)\cdot5.5°=2007.5°\equiv-152.5°=-152°30'.$$
The negative sign is because the hours hand is ahead.
A: Let $\alpha$ be the angle in degrees of the hour hand, measured with reference to $12$ and $\beta$ be the angle in degrees of the minute hand measured with reference to $12$.
At 6:05, the minute hand has moved $\frac{1}{12}$ of the  way around the clock. Thus, $\beta=\frac{360^{\circ}}{12}=30^{\circ}$. 
The hour hand has moved $\frac{1}{12}$ of the way from $6$ to $7$. In other words, $\frac{1}{12}\cdot\frac{1}{12}=\frac{1}{144}$ of the way from the $6$. Thus, $\alpha=180+\frac{360}{144}=182.5^{\circ}$. 
Therefore, the angle between the two hands is $\alpha-\beta=182.5^{\circ}-30^{\circ}=152.5^{\circ}$. 
A: The time in minutes (let's ignore the seconds, as there are none here) is 6*60 + 5 = 365 minutes.
The angle for a hand is 360° * time / (time per 360°)
The angle for the hour hand is 
360° * [365 / (12*60)] = 182.5

The angle for the minute hand is
(360° * 365 / 60) % 360 = 30

The angle between the 2 is 152.5°
A: 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
$$\frac12-\frac1{12}+\frac1{144}=\frac{61}{144}=152.5^\circ$$
A: 
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:
$$\text{S}=\frac{\text{M}^{\circ}}{60\space\text{seconds}}=\left(\frac{1}{120}\right)^{\circ}\text{/}\space\text{second}$$


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