题目内容
在△ABC中,已知tanB=
,cosC=
,AC=3
,求△ABC的面积.
| 3 |
| 1 |
| 3 |
| 6 |
设AB、BC、CA的长分别为c、a、b,tanB=
,得B=60°,sinB=
,cosB=
.
又sinC=
=
,应用正弦定理得c=
=
=8.
∴sinA=sin(B+C)=sinBcosC+cosBsinC=
×
+
×
=
+
.
故所求面积S△ABC=
bcsinA=6
+8
.
| 3 |
| ||
| 2 |
| 1 |
| 2 |
又sinC=
| 1-cos2C |
2
| ||
| 3 |
| bsinC |
| sinB |
3
| ||||
|
∴sinA=sin(B+C)=sinBcosC+cosBsinC=
| ||
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
2
| ||
| 3 |
| ||
| 6 |
| ||
| 3 |
故所求面积S△ABC=
| 1 |
| 2 |
| 2 |
| 3 |
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