Clock Hands, part 2

clock hands word problem
It’s 4:30. The acute angle formed by the big hand and the little hand is 45?. To the nearest tenth, what will that angle be when the time is 4:44? Solve by writing a formula which uses the following variables:
N = new angle
S = starting angle


SOLUTION: In the 14 minutes that transpire between 4:30 and 4:44, the big hand will greatly enlarge the angle as it moves away from the small hand; while the small hand will slightly shrink the angle as it “chases after” the big hand. So the general layout of our equation will look like this:

New angle = Starting angle *plus* Big-hand-movement *minus* Small-hand-movement

As established in Problem 1 (https://classsizeofone.com/clock-hands-math-problem-part-1/), the big hand moves at a rate of 6? per minute, while the small hand moves at .5? per minute. The time is known, and is the same for both hands: 14 minutes. So,

N = S + 6(14) – .5(14)

N = 45 + 84 – 7 = 122. . . . At 4:44, the acute angle formed by the hands will be 122?.