已知△ABC中,sinA(sinB+3cosB)=3sinC. (I)求角A的大小; (II)若BC=3,求△ABC周长的取值范围.
问题描述:
已知△ABC中,sinA(sinB+
cosB)=
3
sinC.
3
(I)求角A的大小;
(II)若BC=3,求△ABC周长的取值范围.
答
(I)A+B+C=π
得sinC=sin(A+B)代入已知条件得sinAsinB=
cosAsinB
3
∵sinB≠0,由此得tanA=
,A=
3
π 3
(II)由上可知:B+C=
,∴C=2π 3
-B2π 3
由正弦定理得:AB+AC=2R(sinB+sinC)=2
(sinB+sin(
3
-B))2π 3
即得:AB+AC=2
(
3
sinB+3 2
cosB)=6sin(B+
3
2
)π 6
∵0<B<
得2π 3
<sin(B+1 2
)≤1π 6
∴3<AB+AC≤6,
∴△ABC周长的取值范围为(6,9]