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I FEEL LIKE THIS Q IS POSTED IN THE FORUM EARLIER BUT CUD NO

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by pops » Mon Mar 29, 2010 9:21 pm
I am not very convinced with the options you gave as all the options are more than 2 pi (which means whole circle)
But here is my approach:
angle (PRO) = 35
since, CR = CP (Radii of same circle) => angle( CRP) = angle (CPR) = 35
hence, angle (PCR)=110
hence, angle (PCO) = 70
hence, angle (CPO) = 70 (alternate angles of parallel lines)

since, CP=CQ (radii) => angle (CPQ)=angle (CQP) = 70
hence, angle (PCQ) = 180 - 70 - 70 = 40
now, 360 --- > 2pi
hence, 40 ----> 2 pi * 40 / 360 = 2 pi / 9
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by thephoenix » Mon Mar 29, 2010 10:29 pm
pops u have made a slight mistake
360=2 pi r
u have missed it
well thanks for your approach
ans is 2pi


is there any other approach
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by solankijignesh » Mon Mar 29, 2010 10:53 pm
i think we can also solve it using inner angles in circle.

angle PCO = 2 x angle PRO = 70
similarly, angle QCR = 70

major arc PQ = 180+70+70 = 320
Minor arc PQ = (40 / 360) * 18 pi = 2pi

is this approach correct?
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by pops » Mon Mar 29, 2010 11:20 pm
thephoenix wrote:pops u have made a slight mistake
360=2 pi r
u have missed it
well thanks for your approach
ans is 2pi


is there any other approach
perfect... my mistake :(

thanks !!
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by eaakbari » Mon Mar 29, 2010 11:49 pm
Since RC = PC
<CPR = <CRP
Hence <p = 70

Alsp <p=<q

Hence <pcq = 180 - 70 - 70 = 40

Hence arc length = 40/360 * 2 *pi*9

Answer = 2pi
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