Seven and a Half Degrees at a Quarter Past Three

Look at an analogue clock at a quarter past three. The minute hand points straight at the 3. The hour hand also looks like it is pointing at the 3. So the angle between them is zero, and the question was a waste of your time.

It is 7.5 degrees. The mistake is not arithmetic and it is not carelessness. It is a picture of a clock that almost nobody examines, and the whole problem lives in that picture rather than in any calculation.

Two bearings, measured the way a dial is read

Fix a convention and the ambiguity disappears. Measure every direction in degrees clockwise from twelve, which is how a dial is read anyway. The minute hand crosses the whole face in sixty minutes, so it moves at $6$ degrees per minute. The hour hand crosses the whole face in twelve hours, so it covers $360/12 = 30$ degrees per hour, which is $0.5$ degrees per minute. At time $H\!:\!M$ the two bearings are therefore

At $H = 3$ and $M = 15$ the minute hand sits at $6 \cdot 15 = 90$ degrees, which is exactly the 3 mark. The hour hand sits at $30 \cdot 3 + \tfrac{1}{2}\cdot 15 = \tfrac{195}{2}$ degrees, or 97.5. The gap is the difference.

In radians that is $\tfrac{15}{2}\cdot \tfrac{\pi}{180} = \tfrac{\pi}{24}$, a number the fifty seconds of video had no room to pronounce and which is worth having written down somewhere. It is a small angle. On a dial 10 centimetres across, 7.5 degrees of rim is about 6.5 millimetres of arc, roughly the width of the numeral printed there, which is why the eye reports zero.

The hour hand crawls, it does not wait

Equation (1) hides the one physical fact the problem is testing, so it is worth stating without symbols. Between 3:00 and 4:00 the hour hand travels from the 3 to the 4. It does not sit on the 3 for an hour and then jump. A quarter of the way through the hour it has covered a quarter of the distance, and that distance is one twelfth of the dial, so a quarter of it is

That is the same 7.5, reached without ever computing a bearing. The minute hand has arrived at the 3 and the hour hand has already left it.

The trap comes from a clock that does not exist

It is tempting to describe the zero answer as a confusion with 3:00, but that is wrong and worth correcting carefully. At 3:00 the minute hand is on the 12 and the hour hand is on the 3, so the angle is a full $90$ degrees. Nothing about 3:00 gives zero.

The zero comes from a clock whose hour hand waits on each numeral until the hour is up and then steps. On that clock the hour hand sits at $30 \cdot 3 = 90$degrees for the whole of the three o’clock hour, and at 3:15 the minute hand is also at 90, so the difference really is $90 - 90 = 0$. The arithmetic is exact. The clock is what is fictional.

This distinction matters more than it looks. A candidate who thinks he confused two times will go back and re-read the question. A candidate who realises he was reading a clock that does not exist has learned something he can reuse, because the same slip produces wrong answers at every other time of day too.

One formula for every time on the dial

Subtracting the two expressions in (1) gives a single closed form. The bearing difference is $30H + \tfrac{1}{2}M - 6M = 30H - \tfrac{11}{2}M$, and since only the size of the gap matters we take the absolute value.

At $H = 3$, $M = 15$ the formula returns $|90 - 82.5| = 7.5$, which is already inside $[0,180]$ and needs no reduction.

The reduction step is not decoration. At 8:00 the raw difference is 240, and no clock ever shows an angle of 240 degrees between its hands, because two rays out of a common point cut the plane into two arcs and the angle between them is the smaller one. So 240 becomes $360 - 240 = 120$.

Where the 5.5 comes from

The coefficient $5.5$ looks arbitrary until you stop tracking two hands and start tracking the gap between them. The minute hand turns at 360 degrees an hour, the hour hand at 30, so the minute hand gains $360 - 30 = 330$ degrees an hour on the hour hand, which is $330/60 = 5.5$ degrees a minute.

Run the clock from 3:00 with that single number. At 3:00 the minute hand is 90 degrees behind the hour hand. Fifteen minutes later it has closed $15 \cdot 5.5 = 82.5$ degrees of that deficit, leaving

Same answer, and this time the picture is a chase rather than two independent positions. The relative view also settles a question the absolute view makes fiddly: the hands coincide when the gap is zero, which happens every $360/330$ hours, so eleven times in twelve hours and not twelve. The missing twelfth coincidence is the one at 12:00, which is shared between the start and the end of the cycle.

What the answer quietly assumes

Equation (1) is a model, and it is worth saying out loud which clocks it describes. Every mechanical movement drives the hour hand off the same train as the minute hand, so it advances smoothly. Most quartz movements do the same for the hour hand even when the second hand ticks. On those clocks 7.5 degrees is exact.

Some movements, and many digital renderings of an analogue face, advance the hour hand in discrete steps. If the step is one whole minute of dial the hour hand has taken fifteen steps of $0.5$ degrees by 3:15 and lands on 97.5 anyway, so the answer survives. If the step is ten minutes it has taken a single step of 5 degrees, landing at 95, and the angle reads $5$ degrees instead. If the hour hand steps only once an hour, the answer really is zero. Any step size that divides fifteen minutes returns 7.5 exactly, and every other step size returns something else. The problem is well posed only once the continuous reading is pinned, which is why it gets pinned here rather than assumed.

Two smaller points close the case. The answer does not depend on 3:15 being a quarter hour in any deep way, since (4) works at every minute and at fractional minutes too. And the angle never exceeds 180 by construction, so there is no time of day at which the formula returns something a dial could not display.

Sources and further reading

• The closed form and its history — Clock angle problem
• The unit the answer is stated in — Degree (angle) and Radian
• The 5.5 degrees a minute of section five — Angular velocity
• Why a mechanical hour hand cannot wait — Movement (clockwork)

Every number here was checked two ways before publication. The bearings and the difference were carried as exact fractions, so 7.5 is $15/2$ and 97.5 is $195/2$ and nothing was ever rounded. Separately, each hand was turned into a unit vector and the angle read off the dot product, a computation that knows no clock formula at all; it agrees with equation (4) at all 720 whole-minute times on a twelve-hour dial, the largest disagreement being about $6 \times 10^{-13}$ degrees of floating-point noise.