Question

已知三角形ABC的面积为S,且S=根号3/4(b²+c²-a²)a=6，求△ABC周长的取值范围

Answer 1

S=√3/4(b²+c²-a²)
又：S=1/2bcsinA
∴√3/4(b²+c²-a²)=1/2bcsinA
又，根据
余弦定理
，b²+c²-a²=2bccosA
∴√3/2
bccosA
=1/2bcsinA
∴tan
=
√3
A
=
π/3
sinA
=
√3/2
B+C
=
π-A
=
2π/3
0＜B，C＜2π/3
C
=
2π/3
-
B
正弦定理
：
b/sinB
=
c/sinC
=
a/sinA
=
6/(√3/2)
=
4√3
b
=
4√3
sinB
，c
=4√3
sinC
a+b+c
=
6
+
4√3
sinB
+
4√3
sinC
=
6
+
4√3｛sinB
+
sinC｝
=
6
+
4√3｛sinB
+
sin(2π/3-B)｝
=
6
+
4√3｛sinB
+
sin2π/3cosB-cos2π/3sinB｝
=
6
+
4√3｛sinB
+
√3/2cosB+1/2sinB｝
=
6
+
4√3｛3/2sinB
+
√3/2cosB｝
=
6
+
12｛√3/2sinB
+
1/2cosB｝
=
6
+
12｛sinBcosπ/6
+
cosBsinπ/6｝
=
6
+
12sin(B+π/6)
0＜B＜2π/3
π/6＜B+π/6＜5π/6
当B+π/6
=
π/2时，有最大值6+12
=
18
当B+π/6趋近π/6或5π/6时，趋近最小值6+12*1/2
=
12
综上，周长
取值范围
：（12，18】