After 9 o'clock,at what time between 9 pm and | vedclass

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(C) The formula for the angle between the hands of a clock is $	heta = |30H - \frac{11}{2}M|$,where $H$ is the hour and $M$ is the minutes.For the ha

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$16 \frac{4}{11}$ minutes past $9$

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(C) The formula for the angle between the hands of a clock is $	heta = |30H - \frac{11}{2}M|$,where $H$ is the hour and $M$ is the minutes.For the hands to point in opposite directions,the angle $	heta$ must be $180^{\circ}$.Given $H = 9$,we have $180 = |30(9) - \frac{11}{2}M|$.$180 = |270 - \frac{11}{2}M|$.Since the minute hand must be past the $9$ position to create a $180^{\circ}$ angle,we solve $270 - \frac{11}{2}M = 180$.$\frac{11}{2}M = 270 - 180 = 90$.$M = \frac{90 	imes 2}{11} = \frac{180}{11} = 16 \frac{4}{11}$ minutes.Therefore,the time is $16 \frac{4}{11}$ minutes past $9$.

After $9$ o'clock,at what time between $9$ pm and $10$ pm will the hour and minute hands of a clock point in opposite direction?

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