Please find the attached file.
Let's say centre of a circle is N.
We are told that <ORP =35
The diameter is a hypothenusis(always), so <OPR =90, hence <POR=180-90-35=55.
Because NO=NP, <OPN=55, too. Hence, <ONP=180-2*55=70.
Because NP and NQ are equal AND OR and PQ are parallel, <QNR=70, too.
minor arc = (40/360)*18Pi= 2pi
By now we are done
GeomQ.doc [23.5 KiB]
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