03 CE Maths Paper 2 Q49 and 50－部落格

標題:

03 CE Maths Paper 2 Q49 and 50

發問:

Give clear solutions about these two questions. Respondent with the most easy-to-follow solutions will be chosen.

最佳解答:

Q49) http://i32.photobucket.com/albums/d50/pzsing/ce0349-1.jpg draw a line from Q to cut SR at T such that PQTS is a parallelogram. ST = PQ = x(property of //gram) ∴TR = SR - ST = y - x ∠RTQ = ∠RSP = α(corr. ∠, PS//QT) ∠TQR = 180?- α - β(∠s sum of △) By sine rule, QT/sinβ = TR/sin(180?- α - β) QT = (y-x)sinβ/sin(180?- (α + β)) QT = (y-x)sinβ/sin(α + β) PS = QT(property of //gram) PS = (y-x)sinβ/sin(α + β)(cm) ANS: D ------------------------------------------ Q50.) http://i32.photobucket.com/albums/d50/pzsing/ce0350.jpg AD is a diameter (given) ∴∠AED = 90?(∠ in semi-circle) ∵AB = BC = CD arc AB = arc BC = arc CD(equal chords, equal arcs) arc AB: arcBD = 1:2 ∠AEB:∠BED = 1:2(angles are proportional arcs) ∵∠AEB + ∠BED = ∠AED = 90? ∠BED = 60? consider the quadrilateral BCDE it is a cyclic quadrilateral ∠BCD + ∠BED = 180?(opp.∠s, cyclic quad.) ∠BCD = 180?- 60? = 120? x + y + z + ∠BCD + ∠AED = (5-2)180? (∠s sum of polygon) x + y + z + 120?+ 90? = 540? x + y + z = 330? ANS: C

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