It’s 3:00 on an analog clock – the big hand is on the “12” and the little hand is on the “3”. The two hands form a 90 degree angle. What size angle will the hands form when the time is 3:10? Write a formula using the following variables, and then solve: S = starting angle, N = new angle.

SOLUTION: When the big hand moves from its current position, it approaches the small hand, shrinking the angle. When the small hand moves, it moves away from the big hand, enlarging the angle. So the general layout of the equation would go like this:

New angle = Starting angle *Minus* Movement-of-big-hand *Plus* Movement-of-small-hand

The hands move at different rates, so let’s figure out each of those rates. Imagine the big hand sitting on the “12” moves 5 minutes so it’s sitting on the “1”. That’s obviously a change of 5 minutes but it’s also a movement of 30?, since the original angle was 90? and moving to the “1” represents 1/3 of the distance between the “12” and the “3”. So, if the ratio is 30? per 5 minutes, that’s a unit rate of 6? per minute. (Another way to get there is this: There are 360? in a circle and 60 minutes on a clock face. 360/60 reduces to a ratio of 6:1.) So if the big hand moves 6? in 1 minute, then in 10 minutes it moves is 6?(10) or 60?. Meanwhile the small hand would take a full hour to go from its current position, “3”, to the next number, “4”, which also represents a movement of 30?. So if something moves 30? in 60 minutes, that means its unit rate would be 30/60 or .5 degrees per minute. If something moves .5 degrees in 1 minute, then in 10 minutes it moves .5?(10), which is 5?. So getting back to the general equation above,

N = S – (big-hand-movement) + (small-hand-movement)

let’s plug in the values for the degrees covered by the two hands, plus 90 for the starting angle which was given:

N = 90 – 60 + 5 … So at 3:10, the angle between the two hands is 35?.

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