急!急! F4 Properties Circles q10

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