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One at 12:00 when both minute hand and hour hand points towards 12 and at 12;30 when minute points at 6 and hour at 12.
Actually, I think it depends on the clock.
If the hour hand is at the same place during the entire hour, then 12:30 is correct.
If the hour hand is moving little by little every minute, then
the hour hand won’t be at the 12 at 12:30. It will be mid-way between 12 and 1.
We want to know when the hour hand and the minute hand are 180 degrees apart.
How fast is the minute hand changing from the 12:00 position?
The minute hand goes a full 360 degrees every 60 minutes.
So we can measure how far its moved as a function of time.
Call that function M(t), for minute hand’s displacement from the 12:00 position.
How much it’s moved depends on time, and similar to distance = rate * time,
we have displacement from 12:00 position = (360 degrees/60 min) * time passed.
So M(t) = 6t. If t = 60, then M(t) = 360 (ie. after one hour, it’s gone full circle).
If t = 30, then M(t) = 180 (ie. after half an hour, it’s down half way around the clock 180 degrees.)
Similarly, we can make a function for how much the hour hand has been displaced from the 12:00 position.
H(t) = rate the hour hand travels * time passed.
The hour hand travels at one-twelfth the speed of the minute hand (it takes 12 hrs to go full circle.)
So H(t) = (30 degrees/60 minutes) * time passed.
Or H(t) = ½ t.
This makes sense, since at t = 60, H(t) = 30 (so it’s gone 30 degrees and reaches the 1:00 position.)
At t = 120, H(t) = 60, reaching the number 2 on the clock.
Remember that M(t) and H(t) measure displacement of the minute and hour hands.
We want them to be separated by 180 degrees.
So now we want M(t) – H(t) = 180 degrees.
So we’d like to solve 6t – ½ t = 180. This gives us t = 32.727272…
So the hour hand and the minute hands will be 180 degrees apart after 32.727272 minutes.
You could generalize this. If you wanted them to be separated by 90 degrees,
that would be solving M(t) – H(t) = 90, which gives us 16.3636…