题目内容
设△ABC的内角A,B,C所对的边分别为a,b,c,(a+b+c)(a-b+c)=ac
(1)求B
(2)若sinAsinC=
,求C
(1)求B
(2)若sinAsinC=
(1)120°
(2)15°或45°
(2)15°或45°
(1)∵(a+b+c)(a-b+c)=ac
∴a2+c2-b2=-ac
由余弦定理知cosB=
=-
∴B=120°
(2)由(1)知A+C=60°
∵cos(A-C)=cosAcosC+sinAsinC
= cosAcosC-sinAsinC+2sinAsinC
=cos(A+C)+2sinAsinC
=
+
=
∴A-C=30°或A-C=-30°,∴C=15°或C=45°
∴a2+c2-b2=-ac
由余弦定理知cosB=
∴B=120°
(2)由(1)知A+C=60°
∵cos(A-C)=cosAcosC+sinAsinC
= cosAcosC-sinAsinC+2sinAsinC
=cos(A+C)+2sinAsinC
=
∴A-C=30°或A-C=-30°,∴C=15°或C=45°
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