The reflex angle between the hands of a clock | vedclass

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Question

(D) The position of the hour hand at $10:25$ is equivalent to $10 + \frac{25}{60} = 10 + \frac{5}{12} = \frac{125}{12}$ hours from $12:00$.The angle t

Vedclass · Accepted Answer

$197 \frac{1}{2}^{\circ}$

Vedclass · Answer

(D) The position of the hour hand at $10:25$ is equivalent to $10 + \frac{25}{60} = 10 + \frac{5}{12} = \frac{125}{12}$ hours from $12:00$.The angle traced by the hour hand in $\frac{125}{12}$ hours is $\left(\frac{360^{\circ}}{12} 	imes \frac{125}{12}ight) = 312.5^{\circ}$.The angle traced by the minute hand in $25$ minutes is $\left(\frac{360^{\circ}}{60} 	imes 25ight) = 150^{\circ}$.The angle between the hands is $|312.5^{\circ} - 150^{\circ}| = 162.5^{\circ}$.The reflex angle is $360^{\circ} - 162.5^{\circ} = 197.5^{\circ}$ or $197 \frac{1}{2}^{\circ}$.

The reflex angle between the hands of a clock at $10:25$ is

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