题目内容
在△ABC中,A,B,C所对的边分别为a,b,c,A=
,(1+
)c=2b,
(1)求C;
(2)若
•
=1+
,求a,b,c.
| π |
| 6 |
| 3 |
(1)求C;
(2)若
| CB |
| CA |
| 3 |
(1)由(1+
)c=2b得
=
+
=
则有
=
=
cotC+
=
+
得cotC=1即C=
、
(2)由
•
=1+
推出abcosC=1+
;而C=
,
即得
ab=1+
,
则有
解得
.
| 3 |
| b |
| c |
| 1 |
| 2 |
| ||
| 2 |
| sinB |
| sinC |
则有
sin(π-
| ||
| sinC |
sin
| ||||
| sinC |
| 1 |
| 2 |
| ||
| 2 |
| 1 |
| 2 |
| ||
| 2 |
得cotC=1即C=
| π |
| 4 |
(2)由
| CB |
| CA |
| 3 |
| 3 |
| π |
| 4 |
即得
| ||
| 2 |
| 3 |
则有
|
|
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