yunxing060081
幼苗
共回答了10个问题采纳率:90% 举报
(1)求∠AEC:
用余弦定理求
设AB=1(因为计算过程都是利用比值,所以设正方形边长为a与边长为1,计算结果等效)
则:
AE=1+√2,∠CAB=45°,
CE^2=AC^2+AE^2-2*AC*AE*Cos(∠CAB)
=(√2)^2+(1+√2)^2-2*(√2)*(1+√2)*cos45°
=2+(1+2√2+2)-(2+2√2)=3
即CE=√3.
同样由余弦定理:
COS∠AEC=(AE^2+CE^2-AC^2)/(2*AE*CE)
=((1+2√2+2)+3-2)/2*(1+√2)*√3)=(2+√2)/(√6+√3)
=((2+√2)(√6-√3))/((√6+√3)(√6-√3))=√6/3
所以角AEC=ACOS(√6/3)≈35.26°
(2)求∠AED:
用正切定理求
设AB=1
则tan(∠AED)=AD/AE=1/(1+√2)=(√2-1)/((1+√2)(√2-1))
=√2-1
所以∠AED=atan(√2-1)=22.5°
计算结束.
1年前
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