题目内容
设△ABC的内角A,B,C的对边分别为a,b,c,且cosA=
,cosB=
,b=3,则c=______.
| 3 |
| 5 |
| 5 |
| 13 |
∵A和B都为三角形的内角,且cosA=
,cosB=
,
∴sinA=
=
,sinB=
=
,
∴sinC=sin(A+B)=sinAcosB+cosAsinB=
×
+
×
=
,
又b=3,
∴由正弦定理
=
得:c=
=
=
.
故答案为:
| 3 |
| 5 |
| 5 |
| 13 |
∴sinA=
| 1-cos2A |
| 4 |
| 5 |
| 1-cos2B |
| 12 |
| 13 |
∴sinC=sin(A+B)=sinAcosB+cosAsinB=
| 4 |
| 5 |
| 5 |
| 13 |
| 3 |
| 5 |
| 12 |
| 13 |
| 56 |
| 65 |
又b=3,
∴由正弦定理
| c |
| sinC |
| b |
| sinB |
| bsinC |
| sinB |
3×
| ||
|
| 14 |
| 5 |
故答案为:
| 14 |
| 5 |
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