急! Properties of Circles 12q11

22a) ∠BCF + 30 = x (Ext. ∠ of △)

∠BCF = x - 30

bi) ∠ECF = 180 - (x - 30) = 210 - x

BCD = 210 - x (Vert. opp. ∠s)

∠BAD = ∠BCF = x - 30 (Ext. ∠, cyclic quad.)

∠BAD + ∠AFC + ∠AEC = ∠ECF

x - 30 + 30 + 20 = 210 - x

2x = 190

x = 95

ii) ∠BAD = 95 - 30 = 65

23a) Side length of square = 4 cm

SQ = 4√2 cm

AQ = 2√2 cm

bi) AB = Sum of radii = 2 + r

ii) BQ = 2√2 - 2 - r

c) Drop perpendicular from B to PQ and let the foot be.

Then r/(2√2 - 2 - r) = sin 45 = 1/√2

r√2 = 2√2 - 2 - r

r(√2 + 1) = 2(√2 - 1)

r = 0.34

24a) ∠OBA = 52 (Alt. ∠s, OC//AB)

∠OAB = 52 since △OAB is isos.

∠BOA = 180 - 52 - 52 = 76

∠OCB = 76/2 = 38

b) ∠OBC = 180 - 52 - 38 = 90

Hence CB is a tangent to the circle at B.