Question 1145926
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Note that there will be two answers.  At some time between 2 and 3 o'clock the minute hand will be 160 degrees ahead of the hour hand; then at some later time the minute hand will be 200 degrees ahead of the hour hand, making the angle between the hands 360-200 = 160 degrees.<br>
Solution 1....<br>
(1) At 2:00, the minute hand is at 12 and the hour hand is at 2; the minute hand is 60 degrees behind the hour hand.<br>
(2) The hour hand moves 0.5 degrees per minute (30 degrees in 60 minutes); the minute hand move 6 degrees per minute (360 degrees in 60 minutes).  So the minute hand moves 5.5 degrees per minute faster than the hour hand.<br>     
(3) For the minute hand to get from 60 degrees behind the hour hand (at 2:00) to 160 degrees ahead of the hour hand, the number of minutes required is<br>
{{{220/5.5 = 40}}}<br>
So the first time between 2 and 3 o'clock when the angle between the hands is 160 degrees is 40 minutes after 2:00, at 2:40.<br>
Note again there will be a second time when the angle between the hands is 160 degrees.  That time is not a "nice" time like 40 minutes; I leave it to you to find it if you need to.  (The way the problem is worded, I assume the "nice" answer is the one you were supposed to get.)<br>
Solution 2....<br>
We can formalize the previous solution algebraically.  At 2:00, the angle the minute hand makes with 12:00 is 0 degrees; the angle the hour hand makes with 12:00 is 60 degrees.  Since the minute hand moves 6 degrees per minute and the hour hand moves 0.5 degrees per minute, we have<br>
60+0.5t = angle the hour hand makes with 12:00 at t minutes after 2:00
0+6t = angle the minute hand makes with 12:00 at t minutes after 2:00<br>
We want the difference between those two angles to be 160 degrees:<br>
{{{(6t)-(60+0.5t)= 160}}}
{{{6t-60-0.5t = 160}}}
{{{5.5t = 220}}}
{{{t = 40}}}<br>
And, as before, we find the angle between the hands is 160 degrees 40 minutes after 2:00, at 2:40.