If the arc length shown is 7.5 feet, find r to the nearest hundredth of a foot.
so i think i can solve this, i just need somebody to double check this for me
x/2 π r = y/360
x = 7.5 i think
The angle is in degree measure.... how troublesome :D
@hero this one doesn't give me the r value, how can i solve without that?
r is what you're supposed to find.
yes so its different than the other ones i have posted
hmm i'm confused
Well, given that your angle is (inconveniently) in degree measure, use this formula for arclength s... \[s = \frac{\theta}{360^o}2\pi r\] Where theta is the angle measure in degrees
ok i see
You already know that s = 7.5, so go all algebra-like on it or somethin' :D
y = 45 x/(2 π r) = 45/360
what does y represent here @hero
arc measure
oh ok
You should have written it in your notes
sorry just got confused for a moment
so i am trying to solve for r, but i need to solve for x first right?
x = arc length which is already given
You have to isolate r
And it won't be easy
would you mind explaining how to isolate r?
\[\frac{x}{2 \pi r} = \frac{y}{360}\] Multiply both sides by r; divide both sides by y \[\frac{x}{2\pi y} = \frac{r}{360}\] Mulitply both sides by 360 \[\frac{360x}{2\pi y} = r\]
and what does x equal again?
i know y=45 but i forgot x
From there, input all the given values x = 7.5 y = 45 \[\frac{360(7.5)}{2 \pi(45)} = r\]
ohh thats right
let me see if i can figure it out from there
r=9.55?
:)
Yes, you got it
yayy, thanks for setting it up for me. i actually understood this one!!
9.55 wouldn't be in degrees would it? is there a specific unit i should use?
You only understand it if you can do it without my help
Like completely on your own. I think you should try doing these first, then asking us to verify your answers.
yes, well i could go back and rework this one because you showed me. the other ones i didn't understand at first but this one was more clear
at first i had no clue how to solve it
well thanks again @hero :) i really appreciate it, i have a few more of these to solve and i'll try to do the rest on my own :)
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