In the figure , ABCD is a cyclic quadrilateral , AC and BD intersect at K and ∠ABD= 50度 . If minor arc AD : minor arc BC=5 : 2 , find ∠AKD
detail step!!!
圖片參考：http://img707.imageshack.us/img707/9607/41172691.png
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2010-10-30 1:10 am
回答 (1)
2010-10-30 1:49 am
✔ 最佳答案
arc AD : arc BC = 5 : 2∠ABD : ∠CAB = 5 : 2 (arc prop. to ∠s at circumference)
Let ∠ABD = 5k and ∠CAB = 2k
∠ABD = 50 deg
5k = 50 deg
k = 10 deg
∠CAB = 2(10 deg) = 20 deg
∠AKD = ∠ABD + ∠CAB (ext. ∠ of Δ)
= 50 deg + 20 deg
= 70 deg
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收錄日期: 2021-04-19 23:28:57
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