標題:
請問呢條數點計?(急!!!!!)[要有詳細步驟]
http://i87.photobucket.com/albums/k138/yamapi_2006/20070306564.jpg 要有詳細步驟!!!!!!!!!!!!!
最佳解答:
a) Since PQ is tangent to the circle, angle PAB = angle ARB = x and angle QAR = angle ABR = y In Triangle ARB, x + y + 90 = 180 x = 90 - y In Triangle QAR, angle ARQ + angle QAR + angle RQA = 180 i.e. angle ARQ + y + 90 = 180 angle ARQ = 90 - y = x = angle ARB = angle ARC ************************ b) Consider Triangle ACR and Triangle AQR, angle ARC = angle ARQ (given from (a)) angle RCA = angle RQA = 90 (given) AR = AR (common side) The two triangles are congr to each other. (AAS) ************************** c) CR = QR (from (b)) Triangle RQC is an iso. triangle, then AR is perpendicular to QC AB II QC ************************* d) AC^2 = 5^2 - 3^2 AC = 4 AQ = 4 Let QR = CR = m, AR^2 = 4^2 + n^2 = n^2 + 16 AR^2 + AB^2 = BR^2 n^2 + 5^2 = (n+3)^2 n^2 + 25 = n^2 + 6n + 9 6n = 16 n = 8/3 = 2.67 Let CQ = 2z z/5 = 2.67/(2.67+3) z = 2.35 CQ = 2z = 4.70 (cm) 2007-03-06 23:05:46 補充： wrong answer in part (d)(AR)^2 (AB)^2 = (BR)^2n^2 16 5^2 = (n 3)^2n^2 41 = n^2 6n 96n = 32n = 16/3 = 5.33Let CQ = 2zz/5 = 5.33/(5.33 3)z = 3.20CQ = 2z = 6.40 (cm)
其他解答:
請問呢條數點計?(急!!!!!)[要有詳細步驟]
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發問:http://i87.photobucket.com/albums/k138/yamapi_2006/20070306564.jpg 要有詳細步驟!!!!!!!!!!!!!
最佳解答:
a) Since PQ is tangent to the circle, angle PAB = angle ARB = x and angle QAR = angle ABR = y In Triangle ARB, x + y + 90 = 180 x = 90 - y In Triangle QAR, angle ARQ + angle QAR + angle RQA = 180 i.e. angle ARQ + y + 90 = 180 angle ARQ = 90 - y = x = angle ARB = angle ARC ************************ b) Consider Triangle ACR and Triangle AQR, angle ARC = angle ARQ (given from (a)) angle RCA = angle RQA = 90 (given) AR = AR (common side) The two triangles are congr to each other. (AAS) ************************** c) CR = QR (from (b)) Triangle RQC is an iso. triangle, then AR is perpendicular to QC AB II QC ************************* d) AC^2 = 5^2 - 3^2 AC = 4 AQ = 4 Let QR = CR = m, AR^2 = 4^2 + n^2 = n^2 + 16 AR^2 + AB^2 = BR^2 n^2 + 5^2 = (n+3)^2 n^2 + 25 = n^2 + 6n + 9 6n = 16 n = 8/3 = 2.67 Let CQ = 2z z/5 = 2.67/(2.67+3) z = 2.35 CQ = 2z = 4.70 (cm) 2007-03-06 23:05:46 補充： wrong answer in part (d)(AR)^2 (AB)^2 = (BR)^2n^2 16 5^2 = (n 3)^2n^2 41 = n^2 6n 96n = 32n = 16/3 = 5.33Let CQ = 2zz/5 = 5.33/(5.33 3)z = 3.20CQ = 2z = 6.40 (cm)
其他解答:
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