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by Ian Stewart » Sun Dec 28, 2008 11:29 am
Seems these triangle/circle relationships are being discussed in a few posts.

There are a few ways to get to an answer here, but in any case, if you need the length of an arc, you need the angle it makes at the center of the circle. So a useful first thing to do here is to connect P and Q to the center (which I'll call A). Now, angle PRO is exactly half of angle PAO (a fact I just proved in a different thread), so PAO is 70 degrees, and by symmetry, QAR is 70 as well. That makes the angle we want, angle PAQ, 40 degrees (because PAO, PAQ and QAR form a straight line and must add to 180). Once you know the angle, the rest of the problem is straightforward.
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by vivek.kapoor83 » Sun Dec 28, 2008 11:37 am
Thanks Ian
but still confused abt how u arrived on 70...i know it s property but not aware fully, if u can throw something on this property and do u hv any idea abt any thread containng all the rules or property of circle.
Thanx in advance.Looking fwd 4 reply.

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by vittalgmat » Mon Dec 29, 2008 7:47 pm
Let me take a crack at this.

Following Ian's convention of A being the center of this circle, there is a rule that the angle at the center of the circle
So I hope you know rule that the angle made at the center of the circle by and arc is 2 times the angle made the same arc at the opposite edge of the circle. (apologies for the terminology error here) So, from this rule,
we get PAO = 70.

Now the chord PQ is parallel to the diameter. So PQ is symmetric.
What do I mean by that??
Draw a diameter such that it is perpendicular to PQ (and OR coz OR || PQ)
Let this line meet PQ at B. Then PB = PQ. This is symmetrical.
Since the figure to the right of this new diameter is a mirror image of figure o the left, we have Angle PAO = QAR.

I hope I have explained correctly..


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by Ian Stewart » Tue Dec 30, 2008 8:43 am
Vivek- I explained why the angle at the center of the circle is double the angle at the endpoint of the diameter here:

www.beatthegmat.com/pls-explain-circle-t26694.html
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