 It is 6:20.

a) What is the acute angle formed by the hands?

b) What will that angle be when the time is 6:54?

SOLUTION:

a) As we established (see Problem 1), the big hand moves at a rate of 6° per minute, while the small hand moves at .5° per minute. Right now the big hand is on the 4 and the small hand is somewhere between the 6 and 7. So, the angle they form is: the angle from 4 to 5, plus the angle from 5 to 6, plus the sliver beyond 6 that the small hand has moved. The wedges between the numbers on an analog clock are all 30° (360°/12 = 30°). So from 4-5 is 30°, and from 5-6 is 30°; we just need to add to this whatever ground was covered by the small hand in the 20 minutes since it was 6:00 (and the small hand was exactly on the 6). If the small hand covers .5°/minute, in 20 minutes it sweeps out an angle of .5°(20) which is 10°. So, adding up our sectors, 30 + 30 + 10 = 70. The current angle formed by the hands is 70°.

b) At 6:54, the big hand will be just below the 11, and the small hand will be just before the 7. Counting up the full sectors, we have three of them: from 7-8, from 8-9, and from 9-10, all 30° each. So the sector from the 7 to the 10 is 90°. We have to add to this the angle between the 10 and the big hand, and the small angle between the 7 and the small hand. If the big hand moves at 6°/minute, then at 54 minutes past the hour it has covered 6•(54) = 324°. We know the gap from the 10 to the 12 is 60°. So think of the 10 as being 300° past the hour (360°-60°). That’s a gap of 324-300 = 24°. Adding that to 90 we are now at 114°. Lastly, if the hour hand moves at .5°/minute, then at 54 minutes past the hour it has moved .5•(54) or 27° past the 6, and is only 3° away from the 7. So the little sliver between the 7 and the hour hand is 3°; adding that to 114 we get 117.

At 6:54, the obtuse angle between the hour hand and the minute hand is 117°.