Two circles intersect in A and B. Quadrilaterals PCBA and ABDE are inscribed in these circles such that PAE and CBD are line segments. If ∠P=95o and ∠C=40o. Also, ∠AED=z. Then the value of z is:
A
65o
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B
105o
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C
95o
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D
85o
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Solution
The correct option is D85o Given−TwocirclesintersectatA&B.ThequadrilateralPCBAisinscribedinonecircleandthequadrilateralEDBAisinscribedinanothercirclesuchthatthepointsP,A&BarecollinearandC,B&Darecollinear.∠AED=z,∠APC=95o&∠PCB=40o.Tofindout−z=?solution−∠APC+∠ABC=180o(sincethesumoftheoppositeanglesofacyclicquadrilateralis180o).⟹∠ABC=180o−∠APC=180o−95o=85o.Again∠ABC+∠ABD=180o(linearpair).∴∠ABD=180o−∠ABC=180o−85o=95o.Now∠ABD+∠AED=180o(sincethesumoftheoppositeanglesofacyclicquadrilateralis180o).∴∠AED=z=180o−∠ABD=180o−95o=85o.Hence,optionDiscorrect.