At what time between 4'o clock and 5'o clock, will the hands of a clock make an angle of $20°$?

Question

Accepted Answer

> [!TLDR]
> Solving $|30H-5.5M|=20^\circ$ for $H=4$ gives two moments in the hour: $4\!:\!18\frac{2}{11}$ and $4\!:\!25\frac{5}{11}$.

> [!FORMULA]
> The angle between the hour hand and minute hand at $H$ hours $M$ minutes (12-hour clock) is
> $$	heta = |30H - 5.5M|\ 	ext{degrees}$$
> since the hour hand moves $0.5^\circ$/min and the minute hand moves $6^\circ$/min, a relative rate of $5.5^\circ$/min.

**Given:** hour $H=4$ (between 4 and 5 o'clock), required angle $	heta = 20^\circ$.

1. Set up the equation:
   $$|30(4) - 5.5M| = 20 \ \Rightarrow\ |120-5.5M| = 20$$
2. **Case 1** (minute hand not yet caught up to $20^\circ$ ahead of the hour mark's angle):
   $$120-5.5M = 20 \Rightarrow 5.5M = 100 \Rightarrow M = \dfrac{100}{5.5} = \dfrac{200}{11} = 18	frac{2}{11}\ 	ext{min}$$
3. **Case 2** (minute hand has moved past, angle now measured the other way):
   $$120-5.5M = -20 \Rightarrow 5.5M = 140 \Rightarrow M = \dfrac{140}{5.5} = \dfrac{280}{11} = 25	frac{5}{11}\ 	ext{min}$$
4. **Self-check** for Case 1: $30(4)-5.5	imes\frac{200}{11} = 120 - \frac{1100}{11} = 120-100=20$. ✓
   For Case 2: $30(4)-5.5	imes\frac{280}{11} = 120-140=-20$, $|{-20}|=20$. ✓

> [!ANSWER]
> Between 4 and 5 o'clock, the hands are $20^\circ$ apart at **$4\!:\!18\frac{2}{11}$** and at **$4\!:\!25\frac{5}{11}$**.

Q.At what time between 4'o clock and 5'o clock, will the hands of a clock make an angle of $20°$ ?

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Q.At what time between 4'o clock and 5'o clock, will the hands of a clock make an angle of 20°20°20°?

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Q.At what time between 4'o clock and 5'o clock, will the hands of a clock make an angle of $20°$ ?

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