Okay, so I need to figure out what time between 5:30 and 6:00 the hands of a clock will be at right angles. Hmm, right angles mean 90 degrees apart. Let me think about how the clock hands move.
First, I recall that the hour hand moves at 0.5 degrees per minute since it takes 12 hours to complete a full circle (360 degrees), so 360/12 = 30 degrees per hour, which is 0.5 degrees per minute. Meanwhile, the minute hand moves faster, right? It completes a full circle in 60 minutes, so that's 360/60 = 6 degrees per minute. So the minute hand moves at 6 degrees per minute.
Since we're looking at the time between 5:30 and 6:00, let's start at 5:30 and see where the hands are at that time. At 5:30, the minute hand is at 6, which is 180 degrees from 12. The hour hand at 5:30 is halfway between 5 and 6. Since each hour is 30 degrees, halfway would be 530 + 15 = 150 + 15 = 165 degrees from 12. Wait, let me check that. Wait, no, at 5:00, the hour hand is at 530 = 150 degrees. Each hour, the hour hand moves 30 degrees, so in 30 minutes, it moves half of that, right? So from 5:00 to 5:30, the hour hand moves 15 degrees, so at 5:30, it's at 150 + 15 = 165 degrees. The minute hand at 30 minutes is 180 degrees.
So the angle between them at 5:30 is 180 - 165 = 15 degrees. But we need a right angle, which is 90 degrees. So now we need to find a time after 5:30 but before 6:00 when the angle between them is 90 degrees.
Wait, but since the minute hand is moving faster than the hour hand, the angle between them increases or decreases? Let me think. At 5:30, the minute hand is at 180 degrees and the hour is at 165. The minute hand is moving at 6 per minute away from 12, and the hour hand is moving at 0.5 per minute. So after 5:30, the minute hand moves faster, so the angle between them would initially increase. Wait, so from 5:30 onward, as the minute hand speeds ahead, the angle between them increases. But once the minute hand passes the hour hand, the angle would start decreasing. Wait, actually, the angle could either be increasing or decreasing, depending on their relative positions. But in this case, maybe between 5:30 and 6:00, how does the angle change?
Alternatively, maybe the equation for the angle between them at any time t minutes past 5:00 would help. Wait, actually, maybe we can model this as a rate problem. Let's think in terms of time after 5:00. So let's let t be the number of minutes after 5:00. Then the angle of the hour hand is 530 + 0.5t = 150 + 0.5t degrees. The angle of the minute hand is 6*t degrees.
But wait, when t is 30 minutes past 5:00, that's 5:30. But we need between 5:30 and 6:00, so t between 30 and 60 minutes.
Wait, perhaps instead of starting from 5:00, maybe starting from 5:30, let's let t be the number of minutes after 5:30. Then, the total time past 5:00 is 30 + t minutes, right? So the hour hand would have moved 150 + 0.5*(30 + t) = 150 + 15 + 0.5t = 165 + 0.5t degrees. The minute hand, starting at 180 degrees (as at 6), then moving t minutes past 5:30 would be 180 + 6t degrees. Wait, but in that case, when t = 0, the minute hand is at 180, hour at 165. So the angle between them is 15 degrees. Then as t increases, the angle increases until and after a point where the minute hand is 90 degrees ahead of the hour hand. Wait, maybe that's when 180 + 6t - (165 + 0.5t) = 90. Wait, this might be a way to model it.
Wait, the minute hand angle minus the hour hand angle would be (180 + 6t) - (165 + 0.5t) = 15 + 5.5t. Setting this equal to 90 degrees, because we want them at a right angle. So, 15 + 5.5t = 90. Then, 5.5t = 75. Therefore, t = 75 / 5.5, which is 750/55, simplifying that gives 150/11 ≈ 13.636 minutes, so approximately 13 and 7/11 minutes. So 5:30 plus 13 and 7/11 minutes is 5:43 and 7/11 minutes.
But wait, another thought: the angle between clock hands can be measured in two ways, the smaller angle and the larger one, so maybe the absolute value of their difference. So perhaps sometimes, the angle could be going past 180. But here, since we are between 5:30 and 6:00, maybe the angle here after 5:30 starts as 15 degrees, then increases up to 15 + 5.5t. Wait, up until when does that equation hold?
Wait, let me think again. Suppose t is the time after 5:30, so starting at t = 0. The minute hand at 180 degrees. Then as t increases, the minute hand moves at 6t degrees from that position, so 180 + 6t. The hour hand moves at 0.5 degrees per minute from 165 degrees (which was its position at 5:30). So at t minutes after 5:30, the hour hand is at 165 + 0.5t. Therefore, the angle between the two hands is |(180 + 6t) - (165 + 0.5t)|.
That's |15 + 5.5t|. Since we're looking for when this equals 90 degrees. So 15 + 5.5t = 90. So solving for t: 5.5t = 75 => t = 75 / 5.5.
But 75 divided by 5.5. Let's compute that. Let's multiply numerator and denominator by 2 to get 150/11, which is approximately 13.636... minutes. Which is 13 minutes and 38.18 seconds. So adding that to 5:30 brings us to 5:43:38.18. So around 5:43 and 38 seconds. But, is there any other time between 5:30 and 6:00 where the angle is 90 degrees?
Wait, maybe. Sometimes, the angle between the clock hands can be 90 degrees twice in a certain interval. For example, between 3:00 and 4:00, they form a right angle twice. But between 5:30 and 6:00, would that happen once or twice? Let's see.
Wait, between 5:30 and 6:00, the minute hand starts at 180 degrees (pointing down) and the hour hand is moving from 5 towards 6. So initially, at 5:30, as previously calculated, the hour hand is at 165 degrees. The angle is 15 degrees. The minute hand moves faster than the hour hand. So if the minute hand starts at 180 degrees (15 degrees ahead of the hour hand). Wait, wait: wait, hour is at 165, minute is at 180, so the angle from hour to minute is 15 degrees, but since the minute is moving away from the hour, increasing the angle. But if the relative speed is 5.5 degrees per minute difference (since the minute moves 6 per min and the hour moves 0.5, the relative speed is 6 - 0.5 = 5.5 per min). So if starting at 15 degrees, how long until that gap is 90 degrees. So that's 75 degrees more, so 75 / 5.5 = 13.636...
But is there another instance where their angle is 90 degrees? For example, when the minute hand laps the hour hand again, but in between 5:30 and 6:00, since the hour hand is at 165 degrees and the minute starts at 180. Then, to get a right angle again, maybe when the minute hand is 90 degrees behind the hour hand. But because the hour hand is moving towards 6, maybe the minute hand would need to be behind. Wait, but the minute hand is moving much faster. Wait, given the progression between 5:30 and 6:00, would the angle ever go beyond 180, and so the smaller angle would be the other way?
Wait for example: at 5:50, let's check manually. The minute hand is at 300 degrees (since 50 minutes is 300 degrees). The hour hand is at 5 + 50/60 hours, so 5.833... hours. So 5.833 * 30 = 175 degrees. Wait, 5.833 x 30, 530 is 150, .83330 is 25, so 175 degrees. So the minute hand is at 300, the hour at 175. So the difference is 300 - 175 = 125 degrees. So the smaller angle would be 125 and the larger 235. So no, not a right angle.
Wait but approaching 6:00. At 6:00, the minute hand is at 0 degrees, hour at 180. Angle between them is 180. So the angle is decreasing from at 5:50 it was 125, then maybe how does it go. Wait, but let's use equations.
Going back, if I model the angle between the two hands as |15 + 5.5t|. When can that be 90 degrees or 270 degrees? But the angle is measured as the smallest angle between the two hands. So maximum angle is 180. Therefore, maybe if the angle is greater than 180, we take 360 minus that. So perhaps, when |15 + 5.5t| = 90, or |15 + 5.5t| = 270. So two solutions. But 270 is over 180, so possibly not considered. So in that case, only one solution where the angle is 90 degrees. So 15+5.5t = 90, which gives t≈13.636 minutes after 5:30. So, 5:30 plus ~13.64 minutes is 5:43 and ~38 seconds. So that would be the single time when they are at right angles.
Alternatively, perhaps an alternative approach. The formula I remember for the angle between two hands is: |30H - 5.5M|, where H is the hour and M is the minutes. Wait, that formula might be useful here. So when the angle between them is 90 degrees, we have |30*5 - 5.5M| = 90. But wait, let me check.
If H is the hour, and M is the minutes, then the hour hand is at 30H + 0.5M degrees, and the minute hand is at 6M degrees. So the angle between them is |30H + 0.5M - 6M| = |30H - 5.5M|. So we set that equal to 90 degrees (or 270, but since we take modulus 360, 270 is equivalent to 90). So to have the angle being right angle, 30H - 5.5M = ±90 + 360k, for some integer k.
Since time is between 5:30 and 6:00, H = 5 and M is between 30 and 60. Hmm, but doesn't H get incremented when M passes 60? But in our case, M is between 30 to 60.
Wait, perhaps we can write H as 5, because between 5:30 and 6:00, so using H = 5. Then we have |305 - 5.5M| = 90. So |150 - 5.5M| = 90. So:
150 - 5.5M = 90 => 5.5M = 150 - 90 = 60 => M = 60 / 5.5 = 10.909... or 150 - 5.5M = -90 => 5.5M = 150 + 90 = 240 => M = 240 / 5.5 = 43.636...
But M is supposed to be between 30 and 60. So M is 43.636... when rounded, that's approximately 43.64 minutes. So 5 hours and 43.64 minutes, which is 5:43 and ~38 seconds. That's within our time frame (between 5:30 and 6:00). The other solution was M = 10.909 minutes which is 5:10 and ~55 seconds, which is before 5:30. So only the 43.636 is applicable here.
Therefore, the time is 5:43 and 43.636 - 43 = 0.636 minutes, which is 38.18 seconds. So approximately 5:43:38.18.
So formal math gives answer to be at M ≈ 43.64 minutes, so 5:43 and 38.18 seconds. So the exact answer is to write the minutes as 150/11.
Wait, because 5.5 in fraction is 11/2, so 240 / (11/2) is 240 * 2 / 11 = 480/11 ≈ 43.636. Hence, 480/11 minutes is the exact value. So the exact time is 5 hours and 480/11 minutes. But converted into minutes and seconds, 480 divided by 11 is 43 with remainder 7, since 11*43=473, so 480-473=7. So 7/11 minutes. Since 1 minute is 60 seconds, so (7/11)*60=420/11 ≈38.18 seconds. So the exact answer is 5:43 and 7/11 minutes, or 5:43: (420/11) seconds.
But the question says "between 5:30 and 6", so hours isn't changing, so the answer is at 5:43 and 7/11 minutes, which can be written as 5:43 and 7/11 minutes or as a fraction, 5 and 43 7/11 minutes.
But maybe the problem expects it written in terms of minutes past 5:00. For example, 5:43 and 7/11 minutes since 5:00 is 43 and 7/11 minutes. But likely, they want the exact time, like H:MM format. So answer is 5:43 and 7/11, which is 5:43:21 9/11, but more standard would be 5:43 and 7/11 minutes, written as a fractional part.
Alternatively, let me check with the original equation. Let’s double-check my first approach where t is minutes after 5:30. So t was 75 / 5.5 = 150/11 ≈13.636 minutes. So 5:30 + 150/11 minutes. 150 divided by 11 is about 13.636, so 5:30 + 13 minutes and 38.18 seconds becomes 5:43 and 38.18 seconds. Yeah, same.
Alternatively, starting from H = 5, M is between 30 and 60. We had |150 - 5.5M| = 90, leading to 150 -5.5M =90 yields M=60/5.5 ≈10.91 (too early) and 150 -5.5M=-90 => M=240/5.5=43.636. Correct.
The formula here is this: For any time H:M, where H is the hour, the angle between the hour and minute hand is |30H -5.5M| degrees. Then they’re at right angles when this equals 90 or 270, but 270 is invalid as it's equivalent to 90. So it's 30H -5.5M = ±90 and then solved with H=5, M between 30 and 60.
Converting 43.636 into minutes, as per 43 and 7/11.
So the answer is at approximately 5:43 and 7/11 minutes.
But to confirm, let me compute it step-by-step for exactness.
Let’s use t as M. So 150 -5.5t = 90. Solving 150 -5.5t =90: subtract 150-90=60, so 5.5t=60. Therefore, t=60/5.5=600/55=120/11≈10.909 minutes, which is too early.
Then 150 - 5.5t= -90 => -5.5t= -240 => t=240/5.5=480/11≈43.636 minutes, which is in the desired range. Therefore, t=480/11 minutes. So converted to minutes and seconds:
480 divided by 11. So 1143 = 473, 480 -473=7. So 7/11 of a minute. Which is 760/11=420/11≈38.1818 seconds. So 5:43:38.1818..., so written as 5:43 and 7/11 minutes.
Therefore, the hands are at right angles at 5:43 and 7/11 minutes. To present this as boxed in the required format, which is the 12-hour format, so \boxed{5:\dfrac{43\frac{7}{11}}} maybe? Wait, fractional minutes are usually expressed as decimals, but in clock problems, fractions are often used.
Alternatively, 480/11 minutes past 5:00. So 5 hours plus 480/11 minutes. But 480/11 minutes is 43+7/11 minutes.
Alternatively, in terms of general formula:
The formula to compute when the minute and hour hands are at right angles is
θ = |30H -5.5M| = 90 (mod 360 is not necessary here). Solve for M when H is known. Since between 5:30 and 6, H is 5 and M is between 30 to 60. So we got M=480/11 ≈43.636. So the exact answer is 43 7/11 minutes past 5:00, which is 5:43 and 7/11.
Therefore, I think the correct answer is \boxed{5:!43!\frac{7}{11}}, but the notation might vary. Alternatively, \boxed{5:\dfrac{43 \frac{7}{11}}}.
But perhaps alternatively:
The problem says "between 5.30 and 6", so starting at 5:30, after how many minutes do they form a right angle. If someone better want to write the exact time as 5 h 43 7/11 min.
Alternatively, sometimes in such problems, the answer is expressed as 32 8/11 minutes past 5, but no, that probably is when they overlap or the first right angle. That may not.
Wait, the key is to get the exact time between 5:30 and 6. So since starting at 5:00, the time is 43 7/11 minutes past 5:00, which is 5:43 and 7/11. So the answer is at 5:43 and 7/11 minutes. So written as 5:43⁷⁄₁₁, boxed. In LaTeX, probably boxed{5:43\tfrac{7}{11}}.
But I should check once again how are these problems usually answered. If possible, check an example. However, in general, use of \dfrac or \tfrac is acceptable for the fraction. Therefore, the final answer is 5:43 and 7/11, which in box form would be \boxed{5:43\tfrac{7}{11}}.
Final Answer The hands of the clock will be at right angles at \boxed{5:43\tfrac{7}{11}}.
Prompt: At what time between 5.30 and 6 will the hands of a clock be at right angles?