Help please....
If the diameter of the circle is 36, what is the length of arc ABC?
A. 8
B. 8pi
C. 28pi
D. 32pi
E. 56pi

- anonymous

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- anonymous

|dw:1442538379561:dw|

- johnweldon1993

Well remember:
\[\large s = r\theta\]
The arc length is equal to the radius of the circle times the angle given in radians
So first, what is the radius? And what is the angle in radians?

- anonymous

18

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## More answers

- anonymous

what do you mean by the angle in radian?

- anonymous

Can you tell me what they mean when they say "An inscribed angle is aways half the central angel."

- johnweldon1993

Okay hang on...lets go one by one lol...So first...the "inscribed angle is always half the central angle"
Lets look at a circle

- johnweldon1993

|dw:1442584881833:dw|

- johnweldon1993

|dw:1442584923535:dw|

- anonymous

ohhhhh

- johnweldon1993

The inscribed angle *According to the Cenntal Angle Theorem* is always equal to HALF the central angle

- anonymous

what does that mean?

- johnweldon1993

SO
If we look again at that formula I gave you
\[\large s = r\theta\]
S = arc length *What we need
r = radius of the circle
\(\large \theta\) = the measure of the central angle in radians

- anonymous

i've never seen that 0 thing before.

- johnweldon1993

\(\large \theta\) = theta = a measure of an angle :)

- anonymous

oh ok

- johnweldon1993

yeah lol sorry :D
Okay so...lets look at your circle
|dw:1442585177036:dw|

- johnweldon1993

Comparing that to my circle I drew above, it looks like you have the INSCRIBED angle labeled here right?
|dw:1442585225131:dw|

- johnweldon1993

SO...since that formula
\[\large s = r\theta\]
Requires the CENTRAL angle...and we have the inscribed...what would be the central angle?

- anonymous

|dw:1442585190505:dw|

- johnweldon1993

Not quite..
So if r = 18 remember
And we have 80 degrees (because remember central is always double the inscribed)
We have
\[\large s = 18 \times (80 \times (\frac{\pi}{180}))\]
\[\large s = 18 \times \frac{4\pi}{9}\]
\[\large s = ?\]
Does that make sense?

- anonymous

the inscribed angle looks bigger than the central angle why is it only half the central angle?

- johnweldon1993

It looks bigger?|dw:1442585541996:dw|

- anonymous

ohhhhh okay i see

- johnweldon1993

Idk if that helped...or made it more confusing...

- johnweldon1993

Okay good lol

- johnweldon1993

So yeah...look back up to my formula up there...Does it make sense?
\[\large s = r\theta\]
\[\large s = 18 \times (80 \times \frac{\pi}{180})\]

- anonymous

yes

- johnweldon1993

Perfect, Also I notice above you had
\[\large \frac{{\pi}}{360}\]
If you want to use 360 on the bottom...make sure you have 2pi on top...because remember 2pi is 360 degrees

- anonymous

2pi is 360 degrees?

- johnweldon1993

Indeed
|dw:1442585940253:dw|

- johnweldon1993

Remember a whole circle is 360 degrees right?

- johnweldon1993

If we travel pi units around the circle...we travel halfway around it...or half of the 360 which is 180

- johnweldon1993

If we then travel another pi units around....2pi....we have traveled around the whole circle...or 360 degrees

- anonymous

ohhhhh

- johnweldon1993

Alright cool, so now that we have that all clear simplify my equation down
\[\large s = 18 \times (\frac{80\pi}{180})\]
\[\large s = 18\times (\frac{4\pi}{9})\]
\[\large s = 2\times 4\pi\]

- anonymous

|dw:1442586054554:dw|

- johnweldon1993

Where is that 36pi term coming from?

- anonymous

|dw:1442586190316:dw|

- anonymous

36 pi is from the diameter

- anonymous

which is the circumference of the whole circle

- anonymous

i subtracted the circumference from the central angle

- anonymous

Is arc ABC the same as arc AC

- johnweldon1993

Well now the only thing is
The want the length of arc ABC
|dw:1442586432497:dw|

- anonymous

|dw:1442586538130:dw|

- johnweldon1993

So we found that length already...the 8pi
By subtracting the 8pi from the whole circumference...we are actually getting
|dw:1442586574680:dw|

- anonymous

Yea i think that's what they wanted

- anonymous

Is there a reason we have to find the central angle to find the arc length? Couldnt we have found it using the inscribed angle?

- anonymous

@johnweldon1993

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