题目内容
在△ABC中,a,b,c分别为三内角A、B、C的对边,
<C<
,0<B<
,且
=
.
(1)判断△ABC的形状;
(2)若|
+
|=2,a=
,求cosB.
| π |
| 3 |
| π |
| 2 |
| π |
| 3 |
| b |
| a-b |
| sin2C |
| sinA-sin2C |
(1)判断△ABC的形状;
(2)若|
| BA |
| BC |
| ||
| 2 |
(1)由
=
及正弦定理,得
=
,
即sinB=sin2C,
∵
<C<
,∴
<2C<π,0<B<
,B+2C=π,
∵A+B+C=π,∴A=C,△ABC为等腰三角形.
(2)由|
+
|=2,得a2+c2+2ac•cosB=4,
∵a=
,∴cosB=
=
=
.
| b |
| a-b |
| sin2C |
| sinA-sin2C |
| sinB |
| sinA-sinB |
| sin2C |
| sinA-sin2C |
即sinB=sin2C,
∵
| π |
| 3 |
| π |
| 2 |
| 2π |
| 3 |
| π |
| 3 |
∵A+B+C=π,∴A=C,△ABC为等腰三角形.
(2)由|
| BA |
| BC |
∵a=
| ||
| 2 |
| 2-a2 |
| a2 |
2-
| ||
|
| 3 |
| 5 |
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