三角形ABC 中, 角C = 90度 , 角A = 45度 , 點D在線段AC上 , 角BDC = 60度 , AD = 1,
求BD ?
請問這數怎麼解
2006-10-30 10:26 pm
回答 (2)
2006-10-31 12:10 am
✔ 最佳答案
From triangle BCD,CD/BD = cos (60 degrees)
CD = 1/2 BD
BC/BD = sin (60 degrees)
BC = sqrt(3)/2 * BD
From triangle ABC,
BC = AC
BC = AD + CD
sqrt(3)/2 * BD = AD + 1/2 * BD
[sqrt(3)-1]/2 *BD = AD
BD = 2AD / [sqrt(3)-1]
BD = 2[sqrt(3)+1]*AD/2
BD = [sqrt(3)+1]*AD
BD = sqrt(3)+1
sqrt(n) = squart root of n
2006-10-30 16:10:51 補充：
square root not squart root
參考: my mathematical knowledge
2006-10-31 1:35 am
三角形ABC,
tan45 = BC / (DC +1)
BC = DC +1 ......(1)
三角形BCD,
tan60 = BC / DC
BC = (「3)(DC) .....(2) [「3 = 開方3 ]
把(1)代入(2), (「3)(DC) = DC +1
DC = 1/ (「3 -1)
三角形BCD,
cos60 = DC / BD
BD = [1/ (「3 -1)] [ 1 / cos60 ]
BD = 「3 + 1
tan45 = BC / (DC +1)
BC = DC +1 ......(1)
三角形BCD,
tan60 = BC / DC
BC = (「3)(DC) .....(2) [「3 = 開方3 ]
把(1)代入(2), (「3)(DC) = DC +1
DC = 1/ (「3 -1)
三角形BCD,
cos60 = DC / BD
BD = [1/ (「3 -1)] [ 1 / cos60 ]
BD = 「3 + 1
參考: MY OWN KNOWLEDGE
收錄日期: 2021-04-12 22:55:45
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https://hk.answers.yahoo.com/question/index?qid=20061030000051KK02629