As follows:
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maths~~~~`
2010-02-01 3:58 am
回答 (2)
2010-02-01 4:12 am
✔ 最佳答案
AB=BC=CD=DE<BCA=<BAC=x(base <s, isos. triangle)
<CBD=2x (ext.<s of triangle)
<CDB=<CBD=2x(base. <s, isos. tirangle)
<BCD=180deg-2(2x) (<s sum of triangle)
<ECD=180deg-[<BCD+<ACB)]
<ECD=180deg-[(180deg-4x)+x] (adj. <s on a st.line)
<ECD=3x
<DEC=<ECD=3x(base. <s, isos. triangle)
y=<CAB + <DEA(ext.<s of triangle)
y=x+3x
4x=y
x=y/4
參考: Hope can help you^^”
2010-02-01 4:07 am
角CAB=角ACB=X
角CBD=角CDB=角CAB+角ACB=2X
角DCE=角CED=角CAB+角CDB=3X
Y=角CAB+角CED=4X
角CBD=角CDB=角CAB+角ACB=2X
角DCE=角CED=角CAB+角CDB=3X
Y=角CAB+角CED=4X
收錄日期: 2021-04-23 20:41:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100131000051KK03480