I need to find the arc of the circle can you please help I have a headache.

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- March 16th 2008, 10:49 AM #1

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- March 16th 2008, 11:11 AM #2
Hello,

I have a headache too

Look at this : Transversal line - Wikipedia, the free encyclopedia (i don't know the name of the angles in english)

Since AB and CE are parallel, angles BAC and ACE are equal.

We know that CE is the bisector of ACD. And ACD is 110°. So what's the value of angle BAC ?

Now, let's calculate angle OAC.

Use the theorem Inscribed angles with the center of the circle in their interior (Inscribed angle - Wikipedia, the free encyclopedia i really can't explain it in English, sorry...)

So you got angle OAC.

Then, you should know that there is a proportionality relation betwen this angle and the length of the arc :

OAC/360 = lengthOfTheArc/(perimeter of the circle)

- March 16th 2008, 11:17 AM #3

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- March 16th 2008, 11:27 AM #4

- March 16th 2008, 11:36 AM #5

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- March 16th 2008, 11:41 AM #6

- March 16th 2008, 11:47 AM #7

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- March 16th 2008, 12:05 PM #8

- March 16th 2008, 12:14 PM #9

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- March 16th 2008, 12:34 PM #10
Ok :/

CE is the bisector of ACD (110°)

So ACE = 110/2 = 55

ACE = BAC, because (CE)//(AB) and are defined by the same line. I don't know the name of the rule in english, i've given you a link to wikipedia that may help you.

So BAC = 55.

We know that BOC, the angle at the center associated with BAC is twice BAC (see the second wikipedia link - i don't know the name of the rule, but we've learnt it in high school in France).

So BOC = 110.

But we know that BC, the arc inscribed by BOC has a length L, given by :

BOC/360 = L/(2 pi R), where R is the radius of the circle.

So L is 110/360 * 2 * pi * R = 11 pi R/18

- March 16th 2008, 12:56 PM #11