標題:

F.4 A.Maths (Triangles)

發問:

免費註冊體驗

 

此文章來自奇摩知識+如有不便請留言告知

1. Show that AC= 開方2 ------------------------------ cos angleC + sin angleC 5. Find EF. http://www.sendspace.com/file/4lcul7 (the given figure)

最佳解答:

(1) Applying sine law: AC/sin 45 = (√2)/sin ∠A AC/sin 45 = (√2)/sin (135 - θ) AC = (√2 sin 45)/sin (135 - θ) = 1/sin (135 - θ) = 1/(sin 135 cos θ - cos 135 sin θ) = 1/[(cos θ)/√2 + (sin θ)/√2] = (√2)/(sin θ + cos θ) (5) By Pyth. theorem, DG = 8/√3 And for the reason of ASA, △DCG and △BAH are congurent. Area of △DCG = △BAH = 4 x 4/√3 x 1/2 = 8/√3 Also for the reason of AAA, △CFG and △CEB are similar with side ratio EB/FG = 3/(4/√3) = (3√3)/4 Now, EB = 8/√3 - EH = 8/√3 - FG EB = 8/√3 - (4/3√3) EB 3√3 EB = 24 - 4 EB (3√3 + 4) EB = 24 Let ∠CFG = ∠CEB = θ, then EB/sin ∠ECB = 3/sin θ sin θ = 3sin ∠ECB / EB = 3 x (4/5) x (3√3 + 4)/24 = (3√3 + 4)/10 With DG//HB: EF sin θ = GB sin ∠GBE EF [(3√3 + 4)/10] = (3 - 4/√3) x (4/5) EF = 8(3 - 4/√3)/(3√3 + 4) = 8(3√3 - 4)/(27 + 4√3)

其他解答:
arrow
arrow
    文章標籤
    酒店 女子 文章 奇摩
    全站熱搜

    香港美食2017 發表在 痞客邦 留言(0) 人氣()