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2019-05-18
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(1)由B=π-(A+C),得(A+C)=π-B,代入2A+C=π/2
得A+π-B=π/2,即B=A+π/2,即sinB=sin(A+π/2),即sinB=cosA,(1)
又因为a/sinA=b/sinB,得a/b=sinA/sinB,把(1)带入得a/b=sinA/cosA=tanA
(2)因为2A+C=π/2,所以0<2A<π/2,即0<A<π/4
由题b/sinB=2sinB/sinB=2,所以a/sinA=c/sinC=2,
a+c=2(sinA+sinC)=2(sinA+sin(π/2-2A))=2(sinA+cos2A)=2(sinA+1-2sin^2A)(0<A<π/4),
解的属于(1,9/4)
得A+π-B=π/2,即B=A+π/2,即sinB=sin(A+π/2),即sinB=cosA,(1)
又因为a/sinA=b/sinB,得a/b=sinA/sinB,把(1)带入得a/b=sinA/cosA=tanA
(2)因为2A+C=π/2,所以0<2A<π/2,即0<A<π/4
由题b/sinB=2sinB/sinB=2,所以a/sinA=c/sinC=2,
a+c=2(sinA+sinC)=2(sinA+sin(π/2-2A))=2(sinA+cos2A)=2(sinA+1-2sin^2A)(0<A<π/4),
解的属于(1,9/4)
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