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F5 Maths
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a) Let O be the center of the circle. ∠TAO = ∠TCO = 90 (tangent⊥radius) ∠AOC = 360 - 90 - 90 - 40 = 140 (∠sum of polygon) ∠ADC = 140/2 = 70 (∠at centre twice ∠ at O ce) (唔識打) b) Join AC. ∠ACD = ∠CAD (equal arcs, equal∠s) So ∠ACD = ∠CAD = (180-70)/2 = 55 (∠sum of △) ∠BCD = 85 (adj. ∠s on st. line) ∠BAD = 95 (opp.∠s, cyclic quad.) So ∠CAB = ∠BAD - ∠CAD = 95 - 55 = 40 In △AOC, OA = OC (radii) ∠OAC = ∠OCA = (180-140)/2 = 20 (∠sum of △) since ∠TAO = 90 (proved) So ∠BAT = ∠TAO - ∠CAB - ∠OAC = 90-40-20=30 2010-10-05 23:24:53 補充： ∠BCD = 85 (adj. ∠s on st. line) ∠BAD = 95 (opp.∠s, cyclic quad.) 可以寫成 ∠DCE = ∠BAD = 95 2010-10-06 18:28:40 補充： actually 證左∠CAB = 40 之後, 大可以咁做 AT = CT (tangent properties) ∠TAC = ∠TCA = 70 (∠ sum of △) ∠BAT = ∠TAC - ∠CAB = 70 - 40 = 30
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F5 Maths
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Figure: http://img227.imageshack.us/img227/1091/95029153.jpg In the figure, TA and TC are tangents to the circle at A and C respectively. BCE is a straight line. Arc AD = arc CD, angle ATC = 40 and angle DCE = 95 a) Find angle ADC. b) Find angle BAT最佳解答:
a) Let O be the center of the circle. ∠TAO = ∠TCO = 90 (tangent⊥radius) ∠AOC = 360 - 90 - 90 - 40 = 140 (∠sum of polygon) ∠ADC = 140/2 = 70 (∠at centre twice ∠ at O ce) (唔識打) b) Join AC. ∠ACD = ∠CAD (equal arcs, equal∠s) So ∠ACD = ∠CAD = (180-70)/2 = 55 (∠sum of △) ∠BCD = 85 (adj. ∠s on st. line) ∠BAD = 95 (opp.∠s, cyclic quad.) So ∠CAB = ∠BAD - ∠CAD = 95 - 55 = 40 In △AOC, OA = OC (radii) ∠OAC = ∠OCA = (180-140)/2 = 20 (∠sum of △) since ∠TAO = 90 (proved) So ∠BAT = ∠TAO - ∠CAB - ∠OAC = 90-40-20=30 2010-10-05 23:24:53 補充： ∠BCD = 85 (adj. ∠s on st. line) ∠BAD = 95 (opp.∠s, cyclic quad.) 可以寫成 ∠DCE = ∠BAD = 95 2010-10-06 18:28:40 補充： actually 證左∠CAB = 40 之後, 大可以咁做 AT = CT (tangent properties) ∠TAC = ∠TCA = 70 (∠ sum of △) ∠BAT = ∠TAC - ∠CAB = 70 - 40 = 30
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