If the length of an arc is 12 inches and the radius of the circle is 10 inches, what is the measure of the arc?
216 degrees
270 degrees
288 degrees

- anonymous

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- schrodinger

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- anonymous

@campbell_st whered that answer go?

- wampominater

do you still need help?

- anonymous

yes please...

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## More answers

- wampominater

ok

- wampominater

so, the first thing you need is the formula for arc length, which is
\[arclength = 2piR(\frac{ \theta }{ 360 })\]

- wampominater

now, look at the values you are given

- wampominater

you have the arc length and the radius, so plug those in.

- wampominater

and what do you get when you do that?

- anonymous

where am i plugging the arc length into?

- wampominater

look at the formula, see where it says arc length? you plug the value you have for arc length right there.

- anonymous

yeah and i can plug in the radius too. but what am i solving for?|dw:1438157454848:dw|

- wampominater

yep! now you want to solve for theta.

- anonymous

theta?

- wampominater

so to make it easier, replace theta with x. it is the symbol in the numerator of the fraction:
\[\theta\]

- anonymous

oh okay.

- wampominater

so the first step is to isolate x, the best way to do that is to first deal with everything outside the parenthesis

- wampominater

you know all the values for those, just solve 2πr

- wampominater

and then, what you want to do is divide both sides by the value that you get for 2πr, which will isolate the fraction and let you solve that next

- anonymous

62.8?

- wampominater

yep!

- wampominater

now divide both sides by 62.8

- wampominater

what do you get?

- anonymous

okay
|dw:1438157770787:dw| now divide by it?

- wampominater

mhm, so it would be 12 over 62.8

- wampominater

that leaves you with
\[\frac{ 12 }{ 62.8 } = \frac{ x }{ 360 }\]

- wampominater

now, go ahead and divide 12 by 62.8, what does that get you?

- anonymous

couldnt you do that as a proportion?

- wampominater

yes, you could if you like

- anonymous

seems easier

- anonymous

|dw:1438157995138:dw|

- wampominater

yep! so that would be the answer i believe.

- anonymous

isnt right

- wampominater

hmm, let me see if i did something wrong

- anonymous

yeah man

- wampominater

hmm, im not sure what I did Wrong, sorry :(

- anonymous

|dw:1438158390561:dw|

- anonymous

@aric200

- UsukiDoll

actually there is a dupe question and that also led to a dead end
http://openstudy.com/study#/updates/52eafc6ae4b0d95ca6b32d90

- wampominater

hmm, could be the question then? idk if you looked at the work usuki, is it right?

- anonymous

its supposed to be 12 pi and copy paste takes out the pi part

- UsukiDoll

I think the question is a bit screwed up... yes ! 12 pi that's what we need

- wampominater

there we go! lol

- wampominater

ok so then, it is 12π and not 12?

- UsukiDoll

because that's also a dupe question
http://openstudy.com/study#/updates/4e2c15f60b8b3d38d3ba2d2d

- anonymous

yep

- wampominater

alright, do you need help still then or are you good?

- anonymous

so its 216?

- UsukiDoll

I don't know if I should touch on this.. if there's a duplicate question. why not look at the past and learn in the future?

- wampominater

good point, i didnt look :p

- wampominater

yes it is 216

- UsukiDoll

I could go over it for verification purposes
so given radius at 10 ( r = 10)
Arc length is 12pi (s=12pi)
we need theta
\[s = \theta r \rightarrow \frac{s}{r} = \theta \]
\[\frac{12 \pi}{10} = \theta \]
that bad boy is in radians and we need degrees so multiply that 12pi/10 with 180/pi so the pi's cancel
\[\frac{12 \pi}{10} \cdot \frac{180}{\pi} = \theta \]
\[\frac{12 }{10} \cdot \frac{180}{1} = \theta \]
\[\frac{2160}{10} = \theta \]
\[216 = \theta \]

- anonymous

what would be the symbol/variable for arc length?

- UsukiDoll

arc length is s
radius is r
angle is the theta

- wampominater

Alright, well thanks for the Help @UsukiDoll , ill be more careful in the future. goodnight!

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