如附圖,每一個小方格均為正方形 若∠ABD=α ∠CBD=β，試求tan(α-β)?
回答 (2)
2016-09-15 10:46 pm
.......E...D
.. |__|__|
A |__|__|
...|__|__|
.............B
α - β = ∠ABE
tan(α - β) = tan∠ABE
tan(α - β) = EA/AB
tan(α - β) = 1
.. |__|__|
A |__|__|
...|__|__|
.............B
α - β = ∠ABE
tan(α - β) = tan∠ABE
tan(α - β) = EA/AB
tan(α - β) = 1
2016-11-12 3:28 am
tanα=1/2
tanβ=1/3
tan(α-β)
=(tanα-tanβ)/(1+tanαtanβ)
=(1/2 -1/3)/[1+(1/2)(1/3)]
=(1/6)/(7/6)
=1/7
tanβ=1/3
tan(α-β)
=(tanα-tanβ)/(1+tanαtanβ)
=(1/2 -1/3)/[1+(1/2)(1/3)]
=(1/6)/(7/6)
=1/7
收錄日期: 2021-04-18 15:33:59
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https://hk.answers.yahoo.com/question/index?qid=20160915055933AASuyfn