题目内容
设△ABC中,cosA=
,sinB=
,则cosC的值为( )
| 3 |
| 5 |
| 5 |
| 13 |
A.
| B.-
| ||||||
C.
| D.
|
由于△ABC中,cosA=
,sinB=
<
,∴sinA=
>
,故A>
,B<
或B>
(舍去).
∴cosB=
,故有cosC=-cos(A+B)=-cosAcosB+sinAsinB=-
×
+
×
=-
,
故选B.
| 3 |
| 5 |
| 5 |
| 13 |
| 1 |
| 2 |
| 4 |
| 5 |
| ||
| 2 |
| π |
| 4 |
| π |
| 6 |
| 5π |
| 6 |
∴cosB=
| 12 |
| 13 |
| 3 |
| 5 |
| 12 |
| 13 |
| 4 |
| 5 |
| 5 |
| 13 |
| 16 |
| 65 |
故选B.
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