Question

第九题，有知道吗？

Answer 1

cos(45度-角BCD)=AC/CD=AC/10. 即有根号2/2 (cosBCD+sinBCD)=AC/10. AC=5根号2(cosBCD+sinBCD).
cos角ABE=cos2倍BCD=AC/BE=AC/(5根号5). AC=5根号5cos2倍BCD.
所以根号5cos2倍BCD=根号2(cosBCD+sinBCD).
根号5cos2倍BCD=根号(cos2倍BCD+1)+根号(1-cos2倍BCD)
5[cos2倍BCD]^2-2=2根号[1-(cos2倍BCD)^2]
25[cos2倍BCD]^4-20[cos2倍BCD]^2+4=4-4(cos2倍BCD)^2
25[cos2倍BCD]^4-16[cos2倍BCD]^2=0.
[cos2倍BCD]^2=0不合理，舍去。
[cos2倍BCD]^2=16/25.
cos2倍BCD=4/5(等于-4/5不合理，舍去)
所以AC=5根号5cos2倍BCD=4根号5.

Answer 2

用co-function property: sin(90-θ) = cosθ 做，感觉会简单一些。
令 θ = ∠BCD
AC = 10cos(45-θ) ... (1)
AC = 5√5cos2θ = 5√5sin(90-2θ) = 10√5sin(45-θ)cos(45-θ) ... (2)
(2)/(1): 1 = √5sin(45-θ) ==> cos(45-θ) = 2/√5
代入 （1），
AC = 20/√5 = 4√5