A 12-hour clock shows that it's m minutes past h hour. What's the angle between the hands?
Let α and β be the respective angles that the long hand and short hand make with the vertical.
We are looking for θ = | α − β |.
Note: For convenience, we'll write the 12th hour as the 0th hour. That is, noon and midnight are 0:00 rather than 12:00.
The fancy terminology for this is modular arithmetic or equivalence classes. But anyone who knows that 12 o'clock plus one hour is usually referred to as 1 o'clock, not 13 o'clock, is doing modular arithmetic.
α = (m/60)(2π) = m(π/30)
β = (h/12)(2π) + (1/12)(m/60)(2π) = h(π/6) + m(π/360)
Therefore, θ = | α − β | = | m/30 − h/6 − m/360 |π = | 11m/360 − h/6 |π
Our result is
θ = (1/2)| 11m − 60h | degrees
or
θ = (π/360)| 11m − 60h | radians
Examples: θ(0:00) = 0 radians
3:00 → π/2 or 90 degrees
3:15 → | 11·15/360 − 1/2 |π = | 11/24 − 1/2 |π = π/24 or 7.5 degrees
3:30 → | 11·30/360 − h/6 |π = | 11/12 − 1/2 |π = 5π/12 or 75 degrees
3:45 → | 11·45/360 − 1/2 |π = | 11/8 − 1/2 |π = 7π/8 or 157.5 degrees
6:00 → π or 180 degrees
9:00 → (3/2)π or 270 degrees
Furthermore, at what times do the hands cross?
They cross whenever θ = 0. So we have
| 11m − 60h |(π/360) = 0
11m = 60h
Let h = 0, then m = 0
h = 1, then m = 60/11 = 5:27 (1:05:27)
h = 2, then m = 120/11 = 10:55 (2:10:55)
h = 3, then m = 180/11 = 16:22 (3:16:22)
...
h = 10, then m = 54:33
Note: The hands cross 11 times every 12 hours.