F.3 Geometry(Topic- Circle)－jpjzhf9的部落格

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F.3 Geometry(Topic: Circle)

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F.3 GeometryTopic: CircleAnswer following question with full reasons.Try to answer the question before 5/2.1. P, Q, R are points on a circle, centre O.∠POQ=54°;∠OQR=36°, and P, R are on opposite sides of OQ.Find ∠QPR and ∠PQR.2. O is the centre of the circle ABC and ∠ABC=30°.If BC is parallel to OA,... 顯示更多 F.3 Geometry Topic: Circle Answer following question with full reasons. Try to answer the question before 5/2. 1. P, Q, R are points on a circle, centre O. ∠POQ=54°;∠OQR=36°, and P, R are on opposite sides of OQ. Find ∠QPR and ∠PQR. 2. O is the centre of the circle ABC and ∠ABC=30°. If BC is parallel to OA, prove that AB⊥OC. *3.Two circles APRB, AQSB intersect at A,B; PAQ, RBS are straight lines. If∠QPR=80° and ∠PRS=70°, find ∠PQS and ∠QSR(hint:join AB)

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I suppose you have drawn the diagrams. Q.1 ∠POQ = 54°(given) ∠PRQ = 54°/2 = 27°(∠ at centre = twice ∠ at circumference) In ΔOPQ, OP = OQ(radii of same circle) ∠OPQ = ∠OQP(base ∠s, isos. Δ) As ∠ sum = 180°, ∠OPQ = ∠OQP = (180°- 54°)/2 = 63° In ΔPQR, ∠PQR = ∠OQP + ∠OQR = 63°+ 36° = 99° ∠QPR + ∠PQR + ∠PRQ = 180°(∠ sum of Δ) i.e. ∠QPR + 99°+ 27° = 180° i.e. ∠QPR = 54° 2007-02-02 21:15:47 補充： Q.2∠ABC = 30°(given)∠AOC = 30°x 2 = 60° (∠ at centre = twice ∠ at circum.)∠OCB = ∠AOC(alt. ∠s, OA // BC) = 60°We call the intersection point of AB and OC as P.In ΔPCB,∠CPB＋∠OCB＋∠ABC = 180° (∠ sum of Δ)i.e. ∠CPB＋ 60°＋30° = 180° i.e. ∠CPB = 90°Hence, AB⊥OC. 2007-02-02 21:25:53 補充： Q.3In cyclic quadrilateral PRBA,∠ABR＋∠APR = 180° (opp. ∠s, cyclic quadrilateral)∠ABR = 180°- 80°= 100°Similarly,∠BAP = 180°-∠BRP = 180°- 70°= 110°For cyclic quadrilateral QSBA,∠PQS = ∠ABR(ext. ∠, cyclic quadrilateral) = 100° 2007-02-02 21:26:12 補充： Similarly,∠QSR = ∠BAP(ext. ∠, cyclic quadrilateral) = 110°--- 全部完成 ---