Question

已知三角形ABC三边a>b>c且a+c=2b,A-C=丌/2,求a:b:c

Answer 1

∵ a+c=2b, A-C=90°,
由正弦定理得
sinA+sinC=2sinB
sinC=sin(A-90° )=-cosA
cosC=cos(A-90°)=sinA
∵ A+B+C=180°
sinA+sinC=2sinB=2sin(A+C)=2sinAcosC+2cosAsinC
sinA+sinC=2sinAcosC+2cosAsinC
sinA+sinC-2sinAcosC-2cosAsinC=0
sinA-cosA -2sinA sinA +2cosAcosA =0
2 sin＾2A+2 sinAcosA –sinA-2 sinAcosA-2cos＾2A+cosA=0
(sinA-cosA)(2sinA+2cosA-1)=0
∵A-C=90°, A=C+90°A≠45°
sinA-cosA=0 不成立。
∴2sinA+2cosA-1=0
sinA+cosA=1/2
sin^2A+cos^2A+2sinAcosA=1/4
2sinAcosA=(1/4)-1=-3/4
2sinAsinC=3/4
sinAsinC=3/8
4sin^2B=(sinA+sinC)^2=sinA^2+sinC^2+3/4=sinA^2+cosA^2+3/4=7/4
sinB=√7/4
sinA+sinC=2sinB
sinA+sinC=√7/2
sinAsinC=3/8
sinA, sinC是方程x^2-√7/2x+3/8=0 两根
sinC=(2√7-1)/8
sinA=√7/2-(2√7-1)/8 =(2√7+1)/8
sinB=√7/4
a:b:c= sinA: sinB: sinC=(2√7+1)/8
：√7/4：(2√7-1)/8=2√7+1：2√7：2√7-1
a:b:c=2√7+1：2√7：2√7-1

Answer 2

解：∵a+c=2b
由正弦定理，得
sinA+sinC=2sinB
即 2sin(A/2+C/2)*cos(A/2-C/2)=2sinB
∴ cos(B/2)*cos(π/4)=2sin(B/2)*cos(B/2)
∴ sin(B/2)=√2/4
则 cos(B/2)=√14/4
sinB=2sin(B/2)*cos(B/2)=√7/4
∴ sinA+sinC=√7/2
又 A-C=π/2
∴ sinA=sin(π/2+C)=cosC
故 cosC+sinC=√7/2
∴ 1+2sinC*cosC=7/4
∴ sin2C=3/4
由a>b>c，A-C=π/2可知
C<π/2
即 2C<π/2
∴ cos2C=√(1-sin²2C)=√7/4
则 cosC==√[(cos2C+1)/2]=(√7+1)/4
即 sinA=(√7+1)/4
故 sinC=√7/2-sinA=(√7-1)/4
∴ sinA：sinB：sinC=(√7+1)/4：√7/4：(√7-1)/4
∵ a：b：c=sinA：sinB：sinC
∴ a：b：c=(√7+1)/4：√7/4：(√7-1)/4