請問如何解這題數學?
回答 (2)
2020-05-11 1:02 pm
✔ 最佳答案
延長CD至F使△FDE為正三角形,由△ABC~△EFC得
AB/BC=EF/FC
2/5=DE/(DE+4)
5DE=2DE+8
DE=8/3
答:(A)
2020-05-11 1:36 pm
因 ∠ACB = ∠DCE, 以 ∠C 表示之.
由三角形餘弦定律:
AC^2 = AB^2+BC^2-2AB．BC．cos60°
= 2^2+5^2-20(1/2) = 19
又:
2^2 = 5^2+19-2(5)√19 cos∠C
故
cos∠C = 4/√19, sin∠C=√3/√19
再由三角形正弦定律:
DE/sin∠C = CE/sin∠D = CD/sin∠E
即
DE/(√3/√19) = CE/sin120° = 4/sin∠E
故
DE = (√3/√19)．4/sin(60°-∠C)
= (4√3/√19)/[(√3/2)cos∠C-(1/2)sin∠C]
= (4√3/√19)/[(√3/2)(4/√19)-(1/2)(√3/√19)]
= 8/3
由三角形餘弦定律:
AC^2 = AB^2+BC^2-2AB．BC．cos60°
= 2^2+5^2-20(1/2) = 19
又:
2^2 = 5^2+19-2(5)√19 cos∠C
故
cos∠C = 4/√19, sin∠C=√3/√19
再由三角形正弦定律:
DE/sin∠C = CE/sin∠D = CD/sin∠E
即
DE/(√3/√19) = CE/sin120° = 4/sin∠E
故
DE = (√3/√19)．4/sin(60°-∠C)
= (4√3/√19)/[(√3/2)cos∠C-(1/2)sin∠C]
= (4√3/√19)/[(√3/2)(4/√19)-(1/2)(√3/√19)]
= 8/3
收錄日期: 2021-04-11 23:07:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200511040214AAz1vrX