Question 1091690
<br>I find it easiest to work problems involving the angle between the hands of a clock by measuring all angles relative to 12 o'clock.<br>
The arithmetic required to solve this kind of problem is based on the fact that the hour hand moves 1/12 of a revolution, or 30 degrees, in 1 hour (60 minutes), so its rate of rotation is 0.5 degrees per minute; and the minute hand moves one full rotation (360 degrees) in an hour (60 minutes), so its rate of rotation is 6 degrees per minute.<br>
You want to find the time between 2 and 3 o'clock when the minute hand is "12 minutes" ahead of the hour hand.  12 minutes is 1/5 of an hour; so you want the minute hand to be 1/5 of 360 degrees, or 72 degrees, ahead of the hour hand.<br>
At x minutes after 2, the angle the hour hand makes with 12 o'clock, in degrees, is
{{{60+0.5x)}}}  (60 degrees to get to 2; then 0.5 degrees for each minute after 2)<br>
And x minutes after 2, the angle the minute hand makes with 12 o'clock, in degrees,  is
{{{6x}}}<br>
You want the minute hand to be "12 minutes", or 72 degrees, ahead of the hour hand, so you want
{{{6x - (60+0.5x) = 72}}}
{{{5.5x = 132}}}
{{{x = 24}}}<br>
The minute hand will be "12 minutes" ahead of the hour hand at 2:24.