Nevernevergiveup wrote:
Dienekes wrote:
gmatprav wrote:
Bingo! central angle theorem is the rock star! Thanks for the valuable Math book Bunuel!
One other thing I missed in this problem is it asks for the minor arc length PQ. NOT length of PQ.
Follow up question. Is it possible to find length of the chord PQ with the information given in the question? If so what is the length of PQ?
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!
DienekesYou don't have to change any value.
I hope below solution helps!
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Another solution not involving trigonometry (if you do know it, youll end up saving a lot of time!).
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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. _________________