c且a+c=2b,A—C=90°,求a:b:c8．在三角形ABC中,若(a+b+c)(a—b+c)=

c且a+c=2b,A—C=90°,求a:b:c8．在三角形ABC中,若(a+b+c)(a—b+c)=

1.a+c=2b=>sinA+sinC=2sinB=>2sin(A+C)/2cos(A-C)/2=4sin(A+C)/2cos(A+C)/2又因0则cos(A-C)/2=2cos(A+C)/2(A-C=π/2)=>cos(A+C)/2=√2/4=>cos(π-B)/2=√2/4sinB/2=√2/4cosB/2=√[1-sin²(B/2)]=√14/4sinB=2(√2/4)(√14/4)=√7/4sinA*sinC=[cos(A-C)-cos(A+C)]/2=[2cos²(A-C)/2-1-2cos²(A+C)/2+1]/2=cos²(A-C)/2-cos²(A+C)/2=(√2/2)²-(√2/4)²=3/8则sinA,sinC是x²-√7x/2+3/8=0的解(又sinA>sinC)sinA=[√7/2+√(7/4-3/2)]/2,sinC=[√7/2-√(7/4-3/2)]/2=>sinA=(√7+1)/4,sinC=(√7-1)/4sinA:sinB:sinC=(√7+1):√7:(√7-1)2.(a+b+c)(a—b+c)=3ac=>(a+c)²-b²=3ac=>a²+c²-b²=accosB=(a²+c²-b²)/2ac=1/2B=π/3tanA+tanC=3+√3=>(sinAcosC+sinCcosA)/cosAcosC=3+√3=>sin(A+C)/cosAcosC=3+√3(A+C=π-B=2π/3)=>cosAcosC=√3/2/(3+√3)=(√3-1)/4又2cosAcosC=cos(A+B)+cos(A-C)=-1/2+cos(A-C)则cos(A-C)=(√3-1)/2+1/2=√3/2因-π则A-C=±π/6若A-C=π/6,A+C=2π/3,得A=5π/12,C=π/4,B=π/3sinA=sin(B+C)=sinBcosC+sinCcosB=(√3/2)(√2/2)+(1/2)(√2/2)=(√6+√2)/4a:b:c=(√6+√2)/4:(√3/2):(√2/2)=(√3+1):√6:2同理:若A-C=-π/6,A+C=2π/3,得C=5π/12,A=π/4,B=π/3a:b:c=2:√6:(√3+1)======以下答案可供参考======供参考答案1:第七题的长度为什么会有度数?

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