题目内容
设△ABC的内角A,B,C的内角对边分别为a,b,c,(a+b+c)(a-b+c)=ac
(I)求B
(II)若sinAsinC=
,求C.
(I)求B
(II)若sinAsinC=
| ||
| 4 |
(I)∵(a+b+c)(a-b+c)=(a+c)2-b2=ac,
∴a2+c2-b2=-ac,
∴cosB=
=-
,
又B为三角形的内角,
则B=120°;
(II)由(I)得:A+C=60°,∵sinAsinC=
,cos(A+C)=
,
∴cos(A-C)=cosAcosC+sinAsinC=cosAcosC-sinAsinC+2sinAsinC=cos(A+C)+2sinAsinC=
+2×
=
,
∴A-C=30°或A-C=-30°,
则C=15°或C=45°.
∴a2+c2-b2=-ac,
∴cosB=
| a2+c2-b2 |
| 2ac |
| 1 |
| 2 |
又B为三角形的内角,
则B=120°;
(II)由(I)得:A+C=60°,∵sinAsinC=
| ||
| 4 |
| 1 |
| 2 |
∴cos(A-C)=cosAcosC+sinAsinC=cosAcosC-sinAsinC+2sinAsinC=cos(A+C)+2sinAsinC=
| 1 |
| 2 |
| ||
| 4 |
| ||
| 2 |
∴A-C=30°或A-C=-30°,
则C=15°或C=45°.
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