高三解三角形
1) 在三角形ABC中, tanB=1,tanC=2 ,b=50 求a 2)在三角形ABC中,A<B<C cot(A/2),cot(B/2),cot(C/2)等差 求cot(A/2)cot(C/2)的值
1) tanA =-tan(B+C)=3 ======>sinA =3√10/10 tanB=1 ====>B=45 ==>sinB=√2/2 正弦定理a=30√5。
2)cot(A/2)+cot(C/2)=2cot(B/2) cos(A/2)/sin(A/2) +cos(C/2)/sin(C/2)=2cos(B/2)/sin(B/2) sin[(A+C)/2]/[sin(A/2)sin(C/2)]=2cos(B/2)/sin(B/2) cos(B/2)/[sin(A/2)sin(C/2)]=2cos(B/2)/sin(B/2) ==>2sin(A/2)sin(C/2)=sin(B/2) =coc[(A+C)/2] 2sin(A/2)sin(C/2)=cos(A/2)cos(C/2)-sin(A/2)sin(C/2) 3sin(A/2)sin(C/2)=cos(A/2)cos(C/2) ==>cot(A/2)cot(C/2)=3 。
1.tanB=1===>sinB=√2/2,cosB=√2/2,
tanC=2===>sinc=√5/5,cosc=2√5/5,
正弦定理c/sinc=b/sinb==>c=bsinc/sinb=10√10
余弦定理b^2=a^2+c^2-2ac cosb
==>50^2=a^2+(10√10)^2-2*a*√2/2
==> a =30√5
2.
1)B=π/4 sinB=√2/2=cosB sinC=2√5/5 cosC=√5/5 a/sinA=b/sinB=c/sinC=2R b/sinB=50√2 tanA+tanB+tanC=tanAtanBtanC 3+tanA=2tanA tanA=3--->sinA=3√10/10 a/sinA=50√2--->a=30√5。
2)cot(B/2)-cot(A/2)=cot(C/2)-cot(B/2) --->2cot(B/2)=cot(A/2)+cot(C/2) --->2tan[(A+C)/2]=cos(A/2)/sin(A/2)+cos(C/2)/sin(C/2) --->2sin[(A+C)/2]/cos[(A+C)/2]=sin[(A+C)/2]/[sin(A/2)sin(C/2)] --->cos[(A+C)/2)]=2sin(A/2)sin(C/2)[sin(A+C)/2<>0] --->cos(A/2)cos(C/2)-sin(A/2)sin(C/2)=2sin(A/2)sin(C/2) --->cos(A/2)cot(C/2)=3sin(A/2)sin(C/2) --->cot(A/2)cot(C/2)=3 。
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