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a_a;" maths 4條 2
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a_a;" maths 4條 2 [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0005-1.jpg[/IMG]
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(9a) ∠COD = 2∠CAD = 100° (Angle at centre = Twice angle at circumference) ∠OEB = ∠OBE (Base angles of isos. triangles) ∠OEB + ∠OBE = ∠COD (Ext. angle of triangle) ∠OEB = 50° ∠BOC = 2∠OEB = 100° (Angle at centre = Twice angle at circumference) (9b) ∠BEP = 90° - ∠BEO since CE is perpendicular to PE ∠BEP = 40° BP = PE (Tangent properties) ∠PBE = ∠PEB = 40° (Base angles of isos. triangle) ∠BPE = 180° - ∠PBE - ∠PEB = 100° (10a) ∠APR = ∠PSR = 50° (Angle in alt. segment) AP = AR (Tangent properties) ∠ARP = ∠APR = 50° (Base angles of isos. triangle) ∠PAR = 80° (10b) AB = AC ∠ABC = ∠ACB = 50° (Base angles of isos. triangle) BQ = BP and CQ = CR (Tangent properties) ∠BQP = ∠BPQ = 65° ∠PSQ = ∠BQP = 65° (Angle in alt. segment) ∠PQS = 180° - ∠PSQ - ∠QPS = 75° (10c) ∠PRS = 180° - ∠PQS (Opp. angles of cyclic quad.) = 105° (11a) Join OO' and O'T, we have: OO' = 16 + r and O'T = r where r is the radius of the smaller circle. Also, draw a perpendicular from O' to OQ and let the foot of perpendicular be D, then O'D = 24 and OD = 16 - r By Pyth. theorem: (16 + r)2 = 242 + (16 - r)2 256 + 32r = 576 + 256 - 32r 64r = 576 r = 9 cm (11b) Join O'C also, we have: O'TC and OQC are similar triangles with side ratio = 16 : 9. Then (24 + TC)/TC = 16/9 24/TC + 1 = 16/9 24/TC = 7/9 TC = 216/7 and OC = 384/7 Finally, area of OQCR is twice the area of OQC, i.e. 384/7 × 24 = 9216/7 cm2. (12a) ∠PQR = 72° ∠RPB = 180° - ∠PRD = 72° since AB//CD So ∠RPB = ∠PQR and hence AB is the tangent to the circle at P for the reason of converse of angle in alt. segment. (12b) ∠PRQ = 72° Then ∠PRE = ∠ERC = 36° ∠REP = ∠RPB = 72° (Angle in alt. segment) So ∠RPE = 180° - ∠REP - ∠ERP = 72° With ∠RPE = ∠REP, REP is an isos. triangle.
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