two circle are intersect at D and C if Angle B A D is equals to 85 degree find X and Y
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Let us denote the centre of C1C1 as O.O.
Draw OQOQ as one of the radii and OMOM as the second radii.
Now as OQOQ is the radii, it is perpendicular to NPQ.NPQ. So, Angle(OQM)=5∘Angle(OQM)=5∘, as by symmetry Angle(MQP)=85∘,Angle(MQP)=85∘, NowAngle(OMQ)=Angle(OQM)Angle(OMQ)=Angle(OQM) as OQ=OMOQ=OM being the same radius,
Now , Angle(MOQ)=170∘Angle(MOQ)=170∘, so angle subtended at the center by the chord MQMQ is 170∘170∘, and Angle(MBQ)=12Angle(MOQ)=85∘.Angle(MBQ)=12Angle(MOQ)=85∘.
Similarly, Angle(A)=Angle(B)=85∘.Angle(A)=Angle(B)=85∘.
which implies Angle(AQB)=10∘.
Draw OQOQ as one of the radii and OMOM as the second radii.
Now as OQOQ is the radii, it is perpendicular to NPQ.NPQ. So, Angle(OQM)=5∘Angle(OQM)=5∘, as by symmetry Angle(MQP)=85∘,Angle(MQP)=85∘, NowAngle(OMQ)=Angle(OQM)Angle(OMQ)=Angle(OQM) as OQ=OMOQ=OM being the same radius,
Now , Angle(MOQ)=170∘Angle(MOQ)=170∘, so angle subtended at the center by the chord MQMQ is 170∘170∘, and Angle(MBQ)=12Angle(MOQ)=85∘.Angle(MBQ)=12Angle(MOQ)=85∘.
Similarly, Angle(A)=Angle(B)=85∘.Angle(A)=Angle(B)=85∘.
which implies Angle(AQB)=10∘.
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