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+

3
cosB)=
3
sinC.
(I)求角A的大小;
(II)若BC=3,求△ABC周长的取值范围.

(I)A+B+C=π
得sinC=sin(A+B)代入已知条件得sinAsinB=

3
cosAsinB
∵sinB≠0,由此得tanA=
3
,A=
π
3

(II)由上可知:B+C=
3
,∴C=
3
-B

由正弦定理得:AB+AC=2R(sinB+sinC)=2
3
(sinB+sin(
3
-B))

即得:AB+AC=2
3
(
3
2
sinB+
3
2
cosB)=6sin(B+
π
6
)

0<B<
3
1
2
<sin(B+
π
6
)≤1

∴3<AB+AC≤6,
∴△ABC周长的取值范围为(6,9]