7.已知三角形ABC的三边a>b>c且a+c=2b,A—C=90°，求a:b:c

Question

7.已知三角形ABC的三边a>b>c且a+c=2b,A—C=90°，求a:b:c

8．在三角形ABC中，若(a+b+c)(a—b+c)=3ac,且tanA+tanC=3+根号3，AB边上的高为4根号3，求角A,B,C的大小与a,b,c的长

有哪位能解这两题的呢？！谢谢了啊！急用！！！

Answer 1

1.a+c=2b
=>sinA+sinC=2sinB
=>2sin(A+C)/2cos(A-C)/2=4sin(A+C)/2cos(A+C)/2
又因0<A/2+C/2<π/2,则sin(A+C)/2≠0
则cos(A-C)/2=2cos(A+C)/2(A-C=π/2)
=>cos(A+C)/2=√2/4
=>cos(π-B)/2=√2/4
sinB/2=√2/4
cosB/2=√[1-sin²(B/2)]=√14/4
sinB=2(√2/4)(√14/4)=√7/4
sinA*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)/4
sinA:sinB:sinC=(√7+1):√7:(√7-1)

2.(a+b+c)(a—b+c)=3ac
=>(a+c)²-b²=3ac
=>a²+c²-b²=ac
cosB=(a²+c²-b²)/2ac=1/2
B=π/3
tanA+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<A-C<π
则A-C=±π/6
若A-C=π/6,A+C=2π/3,得A=5π/12,C=π/4,B=π/3
sinA=sin(B+C)=sinBcosC+sinCcosB
=(√3/2)(√2/2)+(1/2)(√2/2)=(√6+√2)/4
a: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/3
a:b:c=2:√6:(√3+1)

Answer 2

第七题的长度为什么会有度数?