4 Answers By Expert Tutors It helps to create expressions for the angular position of the minute hand and the hour hand, then to set them 90 degrees apart. Hour hand: 360 degrees with 12 hours per circle means 360/12=30 degrees per hour or 0.5 degrees per minute Minute hand: 360 degrees with 60 minutes per circle means 360/60=6 degrees per minuteLet t = time (elapsed since 5:00 in minutes with decimals) H = angle of hour hand (0-360 degrees clockwise) M = angle of minute hand (0-360 degrees clockwise) H = 150 + 0.5t [angle for 5 o’clock plus minutes] M = 0 + 6t [angle for minutes] “perpendicular” means 90 degrees apart (H - M) = 90 or - (H - M) = 90 (1) (H - M) = 90 (150 + 0.5t) - (0 + 6t) = 90 t = 10.909 minutes after 5, or 5:10:55 (hh:mm:ss) or (2) -(H - M) = 90 (0 + 6t) - (150 + 0.5t) = 90 t = 43.636 minutes after 5, or 5:43:38 (hh:mm:ss)
John M. answered • 12/24/16 Engineering manager professional, proficient in all levels of Math
At 5:00 the angle is 150°. Every minute the minute hand moves the angle is decreased by 6° So for every minute the angle decreases, 150 - 6m. Yet the hour hand is moving forward also. It moves one tick, 6...°, for every 12 minutes of the minute hand. So the angle moves at 150 - 6m + 6m/12 For what value of m does that equal 90?
At 5:30 the hands form a 15 degree angle. Min hand moves 6 degrees every min Hour hand moves 1/2 degree every min Therefore, the degree between them changes 5.5 degrees every min. Around 43 and 7/11 minutes after 5
Option 3 : \(7:21\frac{9}{{11}}\)
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Angle made by the hour hand with the minute hand at a given time is \(\frac{{60H - 11M}}{2}\) Where H is the hour and M is the minute. We have H = 7 and M is unknown. Also, the angle should be \(90^\circ \). Therefore, \(90 = \frac{{60 \times 7 - 11M}}{2} \Rightarrow 180 = 420 - 11M \Rightarrow 11M = 420 - 180 = 240\) Thus, we get \(M = \frac{{240}}{{11}} = 21\frac{9}{{11}}\) Hence, the correct answer is \(7:21\frac{9}{{11}}\). India’s #1 Learning Platform Start Complete Exam Preparation
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Oh, let's do this the annoying way. The minute hand moves at $360\frac {\text{degrees}}{\text{revolutions}}\frac {1 \text{revolution}}{60 \text{minutes}}=6\frac{\text{degrees}}{\text{minutes}}$ The hour hand moves at $\frac{360 degrees}{1revolution}\frac{1revolution}{12hours}\frac{1hour}{60 minute}= \frac 12 \frac{degree}{minute}$. If a minute hand is $x$ degrees away from the hour hand how long will it take for the minute hand to become $x + 180$ degrees away from the hour hand? Well that's a matter of solving $x + 6t = x + \frac 12t + 180$ or $t= \frac {180}{\frac {11}{2}}=\frac {360}{11}$. (Roughly $32$ minutes.) At $3:00$ the hour hand and the minute hand are perpendicular. The next time that will happen will be when the minute hand moves for $90$ degrees before the hour had to $90$ degrees after the hour. That will happen in about $32$ minutes. (At $3:32$). Every $\frac {360}{11}$ theminute hand will move 180 degrees further than the hour hand and the hands will be perpendicular. In a $12$ hour (or a $60*12 = 720$ minute period) this will happen $\frac {720}{\frac {360}{11}} = 22$ times. ==== Here's a better way, although it was easy to make annoying sign and logic errors if you aren't careful. The minute hand travels $6$ degrees a minute. ($\frac {360}{60} = 6$). The hour hand travels $.5$ degrees a minute. ($\frac {360}{12*60} = .5$). In a 12-hour period there are $12*60=720$ minutes. Let's say $\theta_t = 6t$ is the angle of the minute hand after $t; 0 \le t < 720$ minutes. Note: it's very possible that $\theta(t) > 360$. Let's say $\phi_t = .5t$ is the angle of the hour hand after $t$ minutes.. Obviously $\theta_t \ge \phi_t$. $\theta_t$ and $\phi_t$ are perpendicular if $\theta_t = \phi_t + 90 + k*180$ for some non-negative integer $k$. So we need the find out how many solutions there are to: $6t = .5t + 90 + k*180; 0 \le t < 720$ there are. So $t = \frac {90 + k*180}{5.5}$ will have one solution for each integer $k$. So how many $k$ are there so that $0 \le t = \frac {90 + k*180}{5.5} < 720$ are there? Multiply everything by $11$: $0 \le 2(90 + k*180)< 11*720$ Divide everything by $180$: $0 \le 1 + 2k < 11*4$ $-1 \le 2k < 11*4 -1$ $\frac 12 \le k < 2*11 - \frac 12$ And as $k$ is a non-negative integer $0 \le k < 22$. There are $22$ possible $k$s which give a solution. This also gives us the times this occurs: $t = \frac {90 + k*180}{5.5} = \frac {180}11 + k *\frac {360}{11} \approx 16.36 + 32.72k$ (or a little more than every half-hour): So at 12:16, 12:49, 1:22, 1:54, 2:27, 3:00, etc. |