7.已知三角形ABC的三边a>b>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)=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
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![7.已知三角形ABC的三边a>b>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)=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](/uploads/image/z/2699271-63-1.jpg?t=7.%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E8%BE%B9a%3Eb%3Ec%E4%B8%94a%2Bc%3D2b%2CA%E2%80%94C%3D90%C2%B0%2C%E6%B1%82a%3Ab%3Ac8%EF%BC%8E%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%8B%A5%28a%2Bb%2Bc%29%28a%E2%80%94b%2Bc%29%3D3ac%2C%E4%B8%94tanA%2BtanC%3D3%2B%E6%A0%B9%E5%8F%B73%2CAB%E8%BE%B9%E4%B8%8A%E7%9A%84%E9%AB%98%E4%B8%BA4%E6%A0%B9%E5%8F%B73%2C%E6%B1%82%E8%A7%92A%2CB%2CC%E7%9A%84%E5%A4%A7%E5%B0%8F%E4%B8%8Ea%2Cb%2Cc%E7%9A%84%E9%95%BF)
7.已知三角形ABC的三边a>b>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)=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
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
7.已知三角形ABC的三边a>b>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)=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
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=√2/4
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
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)
第七题的长度为什么会有度数?