在△ABC中,A、B、C分别为三角形内角,a、b、c为其所对边,已知2√2*(sin^2A-sin^2C)=(a-b)sinB

显示全部楼层 · 2009-8-20 21:27:47

△ABC外接圆半径为√2 R=√2 由正弦定理得 a=2RsinA sinA=a/2√2 sin^2 A=a^2/8 sin^2 C=c^2/8 sinB=b/2√2 2√2(sinA^2-sinC^2)=(a-b)SinB 2√2(a^2-c^2)/8=(a-b)b/2√2 a^2-c^2=ab-b^2 a^2+b^2-c^2=ab 由余弦定理得 cosC=(a^2+b^2-c^2)/2ab=1/2 C=60...

千问 · 2009-8-20 21:27:47

解：△ABC外接圆半径为√2 R=√2 由正弦定理得 a=2RsinA sinA=a/2√2 sin^2 A=a^2/8 sin^2 C=c^2/8 sinB=b/2√2 2√2(sinA^2-sinC^2)=(a-b)SinB 2√2(a^2-c^2)/8=(a-b)b/2√2 a^2-c^2=ab-b^2 ...

千问 · 2009-8-20 21:27:47

由正弦定理a/sinA=b/sinB=c/sinC=2R可将2√2（sin^2A-sin^2C）=（a-b）sinB转化为2√2[(a/2R)^2-(c/2R)^2]=(a-b)b/2R整理得a^2+b^2-c^2=ab∴cosC=（a^2+b^2-c^2）/2ab=1/2 ∴∠C为60° ∴∠A＋∠B=120°S=1/2absi...

千问 · 2009-8-20 21:27:47

(1)△ABC外接圆半径为R=√2.由正弦定理:a/sinA=b/sinB=c/sinC=2R得sinA=a/(2√2),sinB=b/(2√2),sinC=c/(2√2)代入已知条件2√2*(sin^2A-sin^2C)=(a-b)sinB中化简得a2+b2-c2=ab 由余弦定理得 cosC=(a...