(2014•永州三模)在△ABC中,sinA=513,cosB=35,则cosC=(  ) A.-1665 B.-5665 C.±1665 D.±5665

问题描述:

(2014•永州三模)在△ABC中,sinA=

5
13
,cosB=
3
5
,则cosC=(  )
A. -
16
65

B. -
56
65

C. ±
16
65

D. ±
56
65

∵B为三角形的内角,cosB=

3
5
>0,∴B为锐角,
∴sinB=
1−cos2B
=
4
5
,又sinA=
5
13

∴sinB>sinA,可得A为锐角,
∴cosA=
1−sin2A
=
12
13

则cosC=cos[π-(A+B)]=-cos(A+B)=-cosAcosB+sinAsinB=-
12
13
×
3
5
+
5
13
×
4
5
=-
16
65

故选A