Freddie is at chess practice waiting on his opponent's next move. He notices that the 4-inch-long minute hand is rotating around the clock and marking off time like degrees on a unit circle.
Part 1: How many radians does the minute hand move from 3:35 to 3:55? (Hint: Find the number of degrees per minute first.)
Part 2: How far does the tip of the minute hand travel during that time?
Part 3: How many radians on the unit circle would the minute hand travel from 0° if it were to move 3π inches?
Part 4: What is the coordinate points associated with this radian measure?

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- anonymous

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- anonymous

First step is to find how many degrees there are per minute. 60 minutes to 360 degrees. So how many degrees per minute?

- anonymous

6 degrees per minute

- anonymous

Right. And we're trying to figure out how many degrees there are for 20 minutes. \[\large \sf 20 \times 6 =120\] so we're solving for 120 degrees. Do you know what 120 degrees looks like in radian form?

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## More answers

- anonymous

ummmm 2pi/3

- anonymous

Right. So part 1 done. Next it to find the distance the tip traveled. For this we'll use the formula \[\large \sf Arc~=~r \times \theta\]

- anonymous

lol i dont get his part sry

- anonymous

Well wanna use a different formula? We're looking for the distance the tip traveled. In other words, the arc it traveled around the circumference of the clock. |dw:1436398679715:dw|

- anonymous

yes ok

- anonymous

And we'll use the formula \[\large \sf Arc~=~r \times \theta\] to find the arc length.

- anonymous

This is the radian style formula by the way

- anonymous

wait whats the unknowns

- anonymous

So \(\large \sf \theta=\frac{2\pi}{3}\)

- anonymous

yes is r=4?

- anonymous

And r is stated in the top part of your question. Yeah, r is 4

- anonymous

ok so theta= 4pi?

- anonymous

\[\large \sf \theta=\frac{2\pi}{3}\]

- anonymous

i mean arc sorry

- anonymous

\[\large \sf Arc~=~4 \times \frac{2\pi}{3}\]\[\large \sf Arc~=~\frac{8\pi}{3}\]

- anonymous

oh yea sry i multiplied wrong

- anonymous

Anyways, Part 3 uses the same formula. The only difference is what we're solving for. \[\large \sf 3\pi~=~4 \times \theta\]

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