標題:
中4 maths Ex7A 1(b),2(b),4,7,10,15
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中4 maths Ex7A 1(b),2(b),4,7,10,15 http://i164.photobucket.com/albums/u4/ming21ki/IMG_0001-15.jpg [IMG]http://i164.photobucket.com/albums/u4/ming21ki/IMG_0002-14.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/IMG_0006-19.jpg[/IMG]
最佳解答:
1b) c = 35 (alt. angles, AS//DC) ......angleSAT = 90 ( tangent per. radius) ......c + d = 90 ......d = 55 2b) angle OAT = 90 ( tangent per. radius) ......5^2 + 12^2 = OT^2 ......OT=13 ......OB = 5 ( radius) ......therefore y = 8(cm) 4....OT= OB( radiI of the same circle) ......angle OTB = 2f ( base angles, isos. triangle) ......angle OTA = 90 ( tangent per. radius) ......2f + f = 90 ......f = 30 7a).let TA上面個點係 D .....姐係ＴＡＤ係直線 .....angle CAD = 35 ( vert opp. angles) .....angle OAD = 90 ( tangent per. radius) .....therefore angle BAC = 90 - 35 = 55 ..b).angle BCA = 90 (angle in semi-circle) ......angle BCA + CAB + ABC = 180 ( angle sum of triangle ) ......ABC = 180 - 90 - 55 ..............= 35 10a).joinPO ......angle PBO = 30 ( alt angles, CP//BT) ......angleBPO = PBO = 30 (base angles, isos. triangle) ......anglePOT = PBO + BPO = 30+30 = 60( Ext. angle of triangle) ......angle OPT = 90 ( tangent per. radius) ......angle POT(60) + angle OPT(90) + angle PTO = 180 (angle sum of triangle) ......PTO = 30 ...b)angleCPB + BPO + CPA = 90 ......CPB = PBO = 30 (proved) ......CPA = 30 15a)JOIN OT ......OTP = 90 ( tangent per. radius) ......OT^2 + PT^2 = OP^2 ......PT = l/2 ......r^2 + (l/2)^2 = R^2 ( pyth. theo.) ......r^2 = R^2 - (l^2)/4 ......r^2 = [4(R^2) - l/2]/4 .........r = 開方[4(R^2) - l/2]／開方４ .........r = 開方[4(R^2) - l/2]／2 .........r = 1/2 x 開方[4(R^2) - l/2] ....b)OP = 10,PQ = 16 .......therefore PT = 8 .......OT^2 + 8^2 = 10^2 .......OT = 6 .......the radius of the inner circle is 6 cm 2008-01-05 15:09:35 補充： ( tangent per. radius)姐係第２４８頁第一個reference
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