Let $h$ be the hour and $m$ be the minutes.
The hour hand moves $360^\circ$ in 12 hours, or $30^\circ$ per hour, or $0.5^\circ$ per minute.
The minute hand moves $360^\circ$ in 60 minutes, or $6^\circ$ per minute.
We are looking for the time between 5:30 and 6:00, so $h=5$ and $30 \le m < 60$.
The position of the hour hand at time $h:m$ is $30h + 0.5m$ degrees.
The position of the minute hand at time $h:m$ is $6m$ degrees.
We want the hands to be at right angles, so the difference in their positions is $90^\circ$ or $270^\circ$.
Case 1: $6m - (30h + 0.5m) = 90$
$6m - 30(5) - 0.5m = 90$
$5.5m - 150 = 90$
$5.5m = 240$
$m = \frac{240}{5.5} = \frac{2400}{55} = \frac{480}{11} = 43\frac{7}{11}$
So the time is $5:43\frac{7}{11}$. Since $30 \le 43\frac{7}{11} < 60$, this is a valid solution.
Case 2: $6m - (30h + 0.5m) = -90$
$6m - 30(5) - 0.5m = -90$
$5.5m - 150 = -90$
$5.5m = 60$
$m = \frac{60}{5.5} = \frac{600}{55} = \frac{120}{11} = 10\frac{10}{11}$
This time is $5:10\frac{10}{11}$, which is not between 5:30 and 6.
Case 3: $30h + 0.5m - 6m = 90$
$30(5) + 0.5m - 6m = 90$
$150 - 5.5m = 90$
$5.5m = 60$
$m = \frac{60}{5.5} = \frac{600}{55} = \frac{120}{11} = 10\frac{10}{11}$
This time is $5:10\frac{10}{11}$, which is not between 5:30 and 6.
Case 4: $30h + 0.5m - 6m = -90$
$30(5) + 0.5m - 6m = -90$
$150 - 5.5m = -90$
$5.5m = 240$
$m = \frac{240}{5.5} = \frac{2400}{55} = \frac{480}{11} = 43\frac{7}{11}$
So the time is $5:43\frac{7}{11}$. Since $30 \le 43\frac{7}{11} < 60$, this is a valid solution.
So the only time between 5:30 and 6 is $5:43\frac{7}{11}$.
Final Answer: The final answer is $\boxed{5:43:42}$
Prompt: At what time between 5.30 and 6 will the hands of a clock be at right angles?