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12 Oct 2014, 15:04
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bankerboy30
Blue angle is an inscribed angle.
Red angle is a central angle.
The measure of inscribed angle is always half the measure of the central angle.
Check here for more: math-circles-87957.html
11 Feb 2016, 19:59
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gmatprav
Nice question, I am actually trying to figure this. However, I am able to think of a solution for chord length only if the inscribed angle is changed from 35 to 30, making an equilateral triangle.
Bunuel - Can you please help confirm if my understanding is correct, i.e. can we find the exact chord length with the values presented in the question (without changing 35 to 30)?
Thanks!
16 Feb 2016, 08:01
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Dienekes
Dienekes
You don't have to change any value.
I hope below solution helps!
Attachment:
1.png [ 26.87 KiB | Viewed 9464 times ]
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16 Feb 2016, 08:42
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Nevernevergiveup
Another solution not involving trigonometry (if you do know it, youll end up saving a lot of time!).
Attachment:
2-16-16 10-22-54 AM.jpg [ 17.72 KiB | Viewed 9703 times ]
Refer to the image for a description of the points and angles.
Based on your assumption, \(\angle{PRO} = 30\). As \(\angle{PAO}\) is the central angle to\(\angle{PRO}\) , \(\angle{PAO} = 2*30=60\) ---> \(\angle {PAR} = 180-60=120\)
Also, in triangle PAO, PA=OA (as both are the radii of the circle) ---> Triangle PAO is an equilateral triangle --->\(\angle{POA} = 60 = \angle {APO}\)
Now, as OR || PQ --->\(\angle {PRO} = \angle {RPQ} = 30\)
Now consider triangle PAR such that AR=AP=radii ---> (from \(\angle {PAR} = 120\)) you get \(\angle {APR} = \angle {ARP} = 30\)
Finally, in triangle PAQ, \(\angle {APQ} = \angle {APR}+\angle{RPQ} = 30+30 = 60\) and as AP=AQ=radii ---> triangle APQ is another equilateral triangle ---> \(\angle {PAQ} = 60\) or
Alternately, you can see that as PQ||OR ---> \(\angle {PAR }+\angle {APQ} = 180\) (supplementary angles). You have already calculated \(\angle {PAR} = 120\) ---> \(\angle {APQ} = 180-120=60\)
Hence length of PQ = r*theta , where theta = 60 degrees in radians = \(60*2*\pi/360\) =\(\pi / 3\) ---> length minor arc PQ = \(9*\pi/3 = 3\pi\)
Hope this helps.
Do note that the above solution also stands for 35 degrees as I have not used any particular property of equilateral triangles that you couldnt have applied to 35 degrees or any other angle for that matter. I used property of parallel lines, triangles in general and relations of circle and inscribed central angles.
