题目内容
若△ABC的外接圆半径为2,则
+
=______.
| 2b |
| sinB |
| sinC |
| c |
由正弦定理可得
=
=2r=4,
∴
=8,
=
,
∴
+
=8+
=
.
故答案为
.
| b |
| sinB |
| c |
| sinC |
∴
| 2b |
| sinB |
| sinC |
| c |
| 1 |
| 4 |
∴
| 2b |
| sinB |
| sinC |
| c |
| 1 |
| 4 |
| 33 |
| 4 |
故答案为
| 33 |
| 4 |
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