## anonymous one year ago 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?

1. anonymous

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

2. anonymous

6 degrees per minute

3. 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?

4. anonymous

ummmm 2pi/3

5. anonymous

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

6. anonymous

lol i dont get his part sry

7. 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|

8. anonymous

yes ok

9. anonymous

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

10. anonymous

This is the radian style formula by the way

11. anonymous

wait whats the unknowns

12. anonymous

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

13. anonymous

yes is r=4?

14. anonymous

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

15. anonymous

ok so theta= 4pi?

16. anonymous

$\large \sf \theta=\frac{2\pi}{3}$

17. anonymous

i mean arc sorry

18. anonymous

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

19. anonymous

oh yea sry i multiplied wrong

20. 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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