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 \( \square^{\circ} \square^{\prime} \).

**a. Angle swept by the hour hand between noon and \(7{:}15\) P.M.** - The hour hand moves \(30^\circ\) per hour. - In 7 hours it moves: \[ 7 \times 30^\circ = 210^\circ. \] - In 15 minutes (which is \(\frac{15}{60} = \frac{1}{4}\) of an hour) it moves: \[ \frac{1}{4} \times 30^\circ = 7.5^\circ. \] - Total angle swept: \[ 210^\circ + 7.5^\circ = 217.5^\circ. \] - Converting \(0.5^\circ\) to minutes (since \(1^\circ = 60'\)): \[ 0.5^\circ \times 60' = 30'. \] - Therefore, the measure of the angle is \(217^\circ 30'\). **b. Exact angle between the minute and hour hands at \(1{:}25\) P.M.** - **Minute hand:** - It moves \(6^\circ\) per minute. - At 25 minutes past the hour, its position is: \[ 25 \times 6^\circ = 150^\circ. \] - **Hour hand:** - It moves \(30^\circ\) per hour. - At \(1{:}25\) P.M., the hour hand has moved: - \(1\) hour contribution: \[ 1 \times 30^\circ = 30^\circ. \] - Plus additional movement due to 25 minutes: \[ \frac{25}{60} \times 30^\circ = 12.5^\circ. \] - Total position of the hour hand: \[ 30^\circ + 12.5^\circ = 42.5^\circ. \] - **Difference between the hands:** \[ |150^\circ - 42.5^\circ| = 107.5^\circ. \] - Converting \(0.5^\circ\) to minutes: \[ 0.5^\circ \times 60' = 30'. \] - Thus, the exact angle is \(107^\circ 30'\).