maths~~~
2010-04-05 4:37 am
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回答 (1)
2010-04-05 5:33 am
✔ 最佳答案
BC = AB/tan#BD = AB/tan@
CD = AB/tan@ - AB/tan#
2BC = 3CD
2AB/tan# = 3(AB/tan@ - AB/tan#)
2/tan# = 3(1/tan@ - 1/tan#)
2/tan# = 3/tan@ - 3/tan#
5/tan# = 3/tan@
tan@ : tan# = 3 : 5
@ = angleADB
# = angleACB
收錄日期: 2021-04-23 20:39:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100404000051KK01333