圓形角度急急啊！！
2012-01-25 6:50 pm
圖片參考：http://imgcld.yimg.com/8/n/HA05212953/o/701201250013513873424390.jpg
上圖中，弦AC與弦BC交於E。AB=9，BD=15，AD=12，角bac=50度。求角cbd
回答 (4)
2012-01-25 7:07 pm
✔ 最佳答案
先用餘弦定理求角BADcos 角BAD = (9^2 + 12^2 - 15^2)/2 * 9 * 12 = 0
=> 角BAD = 90
因此角CAD = 90 - 50 = 40
而角CBD = 角CAD = 40 (對同弧圓周角)
2012-01-26 1:35 am
角BAE=角ABE=50(等腰三方形性質)
所以
角cbd+50=90(長方形性質)
角cbd=40
所以
角cbd+50=90(長方形性質)
角cbd=40
2012-01-25 7:39 pm
這條應屬幾何問題而非三角問題
AB^2+AD^2=9^2+12^2=81+144
=225=15^2
△BAD是直角三角形
∠BAD=90˚
∠BAC+∠CAD=90˚
50˚+∠CAD=90˚
∠CAD=40˚
∠CBD=∠CAD=40˚
AB^2+AD^2=9^2+12^2=81+144
=225=15^2
△BAD是直角三角形
∠BAD=90˚
∠BAC+∠CAD=90˚
50˚+∠CAD=90˚
∠CAD=40˚
∠CBD=∠CAD=40˚
2012-01-25 7:12 pm
Let angle CBD = x = angle CAD ( angle in the same segment).
By cosine rule:
BD^2 = AB^2 + AD^2 - 2(AB)(AD) cos ( 50 + x)
15^2 = 9^2 + 12^2 - 2(9)(12) cos ( 50 + x)
225 = 81 + 144 - 216 cos(50 + x)
cos (x + 50) = 0
so x + 50 = 90
so x = 40 degree.
By cosine rule:
BD^2 = AB^2 + AD^2 - 2(AB)(AD) cos ( 50 + x)
15^2 = 9^2 + 12^2 - 2(9)(12) cos ( 50 + x)
225 = 81 + 144 - 216 cos(50 + x)
cos (x + 50) = 0
so x + 50 = 90
so x = 40 degree.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120125000051KK00135