急!急! F4 Properties Circles q10－udd18ql48m的部落格

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急!急! F4 Properties Circles q10

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請詳細步驟教我計以下三題 : (不要綱址回答) 圖片參考：http://imgcld.yimg.com/8/n/HA05788109/o/701202270022913873442240.jpg

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28a) ∠QRS = 110 (Opp. ∠s, cyclic quad.) Hence (arc QR + arc RS) : (arc QP + arc PS) = 7 : 11 (2x + 3x) : (5x + arc QP) = 7 : 11 5x/(5x + arc QP) = 7/11 55x = 35x + 7 arc QP arc QP = 20x/7 Hence (arc QP + arc QR) : (arc QR + arc RS) = ∠PSR : ∠QPS 34x/7 : 5x = ∠PSR : 70 ∠PSR = 70 x (34/35) = 68 b) As found in (a), arc QR : arc PQ = 2x : 20x/7 = 7 10 29) DE = x cm (Radius perp. to chord bisects chord) Also ∠ADE = ∠BCE (∠ in the same segment) Hence △BEC ~ △AEC (AAA) x/9 = (x + 4)/x x/9 = 1 + 4/x x2 = 9x + 36 x2 - 9x - 36 = 0 (x - 12)(x + 3) = 0 x = 12 or -3 (rejected) 30) ∠ROS = 180 - 2θ Hence ∠QPS = 180 - 2θ since OR//PQ With ∠QPS should be smaller than 90 (since when ∠QPS = 90, Q merges with P, making a tangent to the circle), we have: 180 - 2θ < 90 θ > 45 Hence 45 < θ < 90