在△ABC中,∠A=
|
由正弦定理可得,
=BC sinA AB sinC
∴sinC=
=ABsinA BC
=
×6 3 2 3 2 2
∵BC>AB
∴
π=A>C1 3
∴C=π 4
∴sinB=sin(A+C)=sinAcosC+sinCcosA
=
×3 2
+2 2
×2 2
=1 2
+2 6 4
故答案为:
,π 4
+6 2 4
在△ABC中,∠A=
|
由正弦定理可得,
=BC sinA AB sinC
∴sinC=
=ABsinA BC
=
×6 3 2 3 2 2
∵BC>AB
∴
π=A>C1 3
∴C=π 4
∴sinB=sin(A+C)=sinAcosC+sinCcosA
=
×3 2
+2 2
×2 2
=1 2
+2 6 4
故答案为:
,π 4
+6 2 4