Question

一道数学题 急~~~ 拜托了啊 20分钟内做出来加分啊

Answer 1

1
BDP相似AEP
角PAE=角B=30，角APE=DPB=(90-30)/2=30
AP=R/tan30=√3,CP=AP/2=√3/2,CE=CPtan30=(√3/2)*(√3/3)=1/2
2
(BD+1)^2-BD^2=3^2 2BD=8,BD=4
sinA=BC/AB=4/5 cosA=3/5,DE^2=2R^2-2R^2cosA=4/5 DE=2√5/5
DE/sinA=1/sinAED=1/sinECP, sinECP=sinA/DE=(4/5)/(2√5/5)=2√5/5
cosECP=(√5/5) tanBPD=cotECP=cosECP/sinECP=1/2
3
tanBPD=1/3, cotAED=1/3 (sinAED)^2=1/[1+(cotAED)^2]=9/13 (cosAED)^2=4/13
sinAED=3/√13,cosAED=2/√13, sinA=sin(180-A)=sin2(AED)=2sinAEDcosAED=12/13
cosA=5/13
CE=x,AC=1+x AB=AC/cosA=(1+x)/(5/13) BC=ABsinA=(1+x)*12/5
y=(1+x)+13(1+x)/5+12(1+x)/5=6(1+x)

Answer 2

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Answer 3

1
BDP相似AEP
角PAE=角B=30，角APE=DPB=(90-30)/2=30
AP=R/tan30=√3,CP=AP/2=√3/2,CE=CPtan30=(√3/2)*(√3/3)=1/2
2
(BD+1)^2-BD^2=3^2 2BD=8,BD=4
sinA=BC/AB=4/5 cosA=3/5,DE^2=2R^2-2R^2cosA=4/5 DE=2√5/5
DE/sinA=1/sinAED=1/sinECP, sinECP=sinA/DE=(4/5)/(2√5/5)=2√5/5
cosECP=(√5/5) tanBPD=cotECP=cosECP/sinECP=1/2