Question
Consider a simple clock. a. Find the measure of the angle swept by the hour hand between noon and \( 7: 15{ }^{\circ} \) P.M. b. Find the exact angle between the minute and hour hands at \( 1: 25 \) P.M. a. The measure of the angle is \( \square^{\circ} \square^{\prime} \).
Ask by John Cox.
Mar 22,2025 03:15
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Answer
**a. Angle swept by the hour hand between noon and \(7{:}15\) P.M. is \(217^\circ 30'\).**
**b. Exact angle between the minute and hour hands at \(1{:}25\) P.M. is \(107^\circ 30'\).
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To find the angle swept by the hour hand between noon and 7:15, we first note that the hour hand moves at a rate of \( 30^\circ \) per hour (since \( 360^\circ \div 12 \)). From noon to 7:00 P.M. is 7 hours, which gives us \( 7 \times 30 = 210^\circ \).
From 7:00 P.M. to 7:15 P.M. is an additional 15 minutes, during which the hour hand moves another \( 7.5^\circ \) (since \( 30^\circ \div 60 \times 15 = 7.5^\circ \)).
Adding these together, the total angle swept is \( 210^\circ + 7.5^\circ = 217.5^\circ \). Therefore, the measure of the angle is \( 217^{\circ} 30^{\prime} \).
For the exact angle between the minute and hour hands at 1:25 P.M., we first calculate the positions of both hands. The hour hand is \( 30^\circ \times 1 + \frac{25}{60} \times 30^\circ = 30 + 12.5 = 42.5^\circ \). The minute hand is at \( 6 \times 25 = 150^\circ \).
The difference in their positions gives us \( |150^\circ - 42.5^\circ| = 107.5^\circ \). Thus, the exact angle between the minute and hour hands at 1:25 P.M. is \( 107.5^\circ \).
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